10 worked examples24 exercises6 figuresbölüm 6 / 14
06Mean Value Theorem, derivatives and graph shape
Sample a function at $x=-2,\;0,\;2$ and read off $-2,\;0,\;2$: three heights climbing steadily left to right. Draw the smooth curve through them and you draw something that rises the whole way. The graph those three points came from drops four units in the middle, and no number of extra samples can ever prove that a drop like that is not hiding between two of them.
By the end of this section you can name every interval where that function rises, falls, bends up and bends down, say where its peaks and valleys are, say what it does far out at both ends, and draw the whole graph — from two and two limits, without plotting a single extra point.
In 60 seconds
The sign of $f'$ decides which way the graph goes, the sign of $f''$ decides which way it bends, and the limits as $x\to\pm\infty$ decide where it ends up — the whole shape is three sign facts collected in a fixed order.
Increasing/Decreasing Test
$f'>0 \text{ on } I \Rightarrow f \text{ increases on } I;\quad f'\lt 0 \Rightarrow f \text{ decreases}$
any question asking for intervals of increase or decrease
$\lim_{x\to\pm\infty} f(x)=L \;\Rightarrow\; y=L \text{ is a horizontal asymptote}$
the two far ends of a sketch; rational and root expressions
Three most common mistakes
Treating every critical number as a peak or a valley. $f(x)=x^{3}$ has $f'(0)=0$ and no extremum at all — the sign of $f'$ has to change.
Announcing an at a number that is not in the domain. For $f(x)=x^{2}/(x^{2}-1)$ the bending does flip across $x=1$, but there is no point of the graph there to inflect.
Reading the sign of $f'$ off the values of $f$. At $x=-0.5$ the function $x^{3}-3x$ is positive and falling at the same time; only $f'(-0.5)=-2.25$ decides the direction.
Midterm 1, Midterm 2 and the Final carry 28 percent each, quizzes 10 and homework 6. What this section produces are interval answers, so the sign line with the test values you actually evaluated is part of the answer, not scratch work.
How much time do you have?
10 minutes
You leave able to build one sign line and classify a critical number — the two moves that show up in every question of this type.
In 60 seconds card, What one derivative sign does to a whole interval, Reading a peak or a valley off the sign change, Building a sign line, Formula card
45 minutes
Add bending, the silent case of the second derivative test, and the two ends of the picture; you can now answer a full 'analyse and sketch' question.
everything in the 10 minute path, Bending: what the sign of the second derivative adds, The one line shortcut, and when it is silent, Far out: horizontal asymptotes, Full exam style question, Practice C
full read
The checklist, the scaffolded ladder, and the interleaved set where the type of the question is hidden — this is the part that transfers to a question you have not seen.
all blocks in order, Putting it together: the checklist, Scaffolding comes off, Practice A to D, Mistake ledger
By the end of this section
Decide from the sign of $f'$ whether $f$ rises or falls on an interval, and say which theorem licenses the jump from one derivative value to a whole interval.
Classify every critical number of a function as a local maximum, a local minimum or neither, and report the value as well as the location.
Determine the intervals of concavity from the sign of $f''$ and locate the inflection points, rejecting the zeros of $f''$ that are not inflection points.
Apply the second derivative test where it works, and recognise the case where it says nothing and the sign of $f'$ must finish the job.
Compute limits at infinity of rational and root expressions by dividing through by the highest power, and turn the results into horizontal asymptotes.
Produce a full sketch of a rational function by running the checklist in order: domain, intercepts, symmetry, asymptotes, $f'$, $f''$.
Syllabus coverage
3.3
What the derivatives say about the shape of a graph
Rising and falling, the first derivative test, concavity, inflection points and the second derivative test are the first four blocks of this section.
covered
3.4
Limits at infinity and horizontal asymptotes
Rational and root expressions are handled by dividing by the highest power; the slanted case is named where it belongs, in the same block.
covered
3.5
Summary of curve sketching
The checklist plus one function carried all the way to a finished picture.
covered
l'Hôpital's rule
The general rule for 0/0 and infinity over infinity limits
Not part of this week: it arrives with the transcendental functions later in the term. Every here is settled by dividing through by the highest power, which is what this week's material asks for.
deferred
Recall first
Critical number
A number $c$ in the domain of $f$ with $f'(c)=0$ or with $f'(c)$ undefined.
These are the breakpoints of every sign line here. The sign of $f'$ can only flip at a critical number or at a number missing from the domain.
Fermat's theorem
If $f$ has a local maximum or minimum at an interior point $c$ and $f'(c)$ exists, then $f'(c)=0$.
It is what makes the search finite: extrema can only hide at critical numbers, so we test those and nothing else.
Mean Value Theorem
If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there is a $c$ in $(a,b)$ with $$f'(c)=\frac{f(b)-f(a)}{b-a}.$$
This is the engine of the whole section: it is the only reason a fact about the derivative at unknown single points controls the function on a whole interval.
Every rational function in this section needs it twice: once for $f'$, once again for $f''$.
Vertical asymptote
If $\lim_{x\to a^{-}}f(x)$ or $\lim_{x\to a^{+}}f(x)$ is $+\infty$ or $-\infty$, the line $x=a$ is a vertical asymptote.
A vertical asymptote is a hole in the domain, so it cuts every sign line in two and no interval statement may cross it.
Closed Interval Method
On a closed bounded interval $[a,b]$ a continuous $f$ attains an absolute maximum and minimum; compare $f$ at the critical numbers inside with $f(a)$ and $f(b)$.
Local and absolute are different questions, and the interleaved practice deliberately mixes them.
Try it yourself first (3 questions)
1§06.0 — what a horizontal tangent does and does not tell you●●○○○
A classmate differentiates, finds $f'(2)=0$ and writes "so $f$ has a minimum at $x=2$". Nothing in this short set is graded and nothing in it is new: each one is a tool the section is built on.
Given
$f$ is differentiable everywhere and $f'(2)=0$.
Find
Which claim about the graph of $f$ at $x=2$ does $f'(2)=0$ justify on its own?
Hint 1/4
Do not compute anything. Read the equation literally: what does the number $f'(2)$ measure on the graph?
Hint 2/4
The derivative at a point is the slope of the tangent line at that point, so $f'(2)=0$ is a statement about one slope.
Hint 3/4
Given data again: $f'(2)=0$, and nothing else about $f$. A slope of zero at a single point means the tangent line there is level.
Hint 4/4
The safe reading is the literal one: the tangent at $x=2$ is horizontal, and that is all.
Show solutionTranslate the symbol
$f'(2)=\text{slope of the tangent at } x=2$
the derivative at a point is defined as that slope, so this is a translation, not a deduction
$f'(2)=0 \Rightarrow \text{tangent at } x=2 \text{ is horizontal}$
zero slope is a level line
Test the tempting extra claim
$f(x)=x^{3}\ \Rightarrow\ f'(0)=0$
one example is enough to kill a general claim
$x^{3} \text{ has no extremum at } 0$
it keeps increasing through the flat moment, so a horizontal tangent alone cannot mean extremum
Answer $$\text{Only: the tangent at } x=2 \text{ is horizontal.}$$
Check
Turn it around: if a horizontal tangent forced an extremum, $x^{3}$ would have one at the origin, and its values $-1,0,1$ at $x=-1,0,1$ say otherwise.
A critical number is a candidate. Every classification in this section is the work of turning a candidate into a verdict.
2§06.0 — first and second derivative in factored form●●○○○
Everything in this section is read off the signs of $f'$ and $f''$, and signs are readable only when the derivative is factored. This is the warm-up for the function used twice later on.
Given
$f(x)=x^{4}-4x^{3}$
Find
Compute $f'(x)$ and $f''(x)$, and write both in factored form.
Hint 1/4
You are not asked where anything is positive yet — only to differentiate twice and factor.
Hint 2/4
Power rule term by term: $\frac{d}{dx}x^{n}=nx^{n-1}$; then repeat on the result.
Hint 3/4
With $f(x)=x^{4}-4x^{3}$: the first derivative is $4x^{3}-12x^{2}$, and the common factor is $4x^{2}$.
Hint 4/4
Factored: $f'(x)=4x^{2}(x-3)$ and $f''(x)=12x(x-2)$.
Show solutionDifferentiate twice
$f'(x)=4x^{3}-12x^{2}$
power rule on each term
$f''(x)=12x^{2}-24x$
power rule again, on f' this time
Factor so the signs are visible
$f'(x)=4x^{2}(x-3)$
the sign of a product is readable; the sign of a difference of powers is not
$f''(x)=12x(x-2)$
same reason, and it exposes the two candidate breakpoints 0 and 2
Check one value in the unfactored and the factored form: at $x=1$, $4-12=-8$ and $4(1)(1-3)=-8$ agree.
Factored derivatives are not cosmetics here; the sign line cannot be built from an unfactored polynomial.
3§06.0 — quotient rule on the function used for the final sketch●●●○○
The last block of this section sketches one rational function from scratch. Its first derivative is the price of admission, and the quotient rule is the only tool needed.
Given
$f(x)=\dfrac{x^{2}}{x^{2}-1}$
Find
Compute $f'(x)$ and simplify it as far as it will go.
Hint 1/4
This is a single application of one rule; no sign analysis is wanted yet.
Hint 2/4
Quotient rule: $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$ with $u=x^{2}$ and $v=x^{2}-1$.
Hint 3/4
With $u=x^{2},\;u'=2x,\;v=x^{2}-1,\;v'=2x$ the numerator is $2x(x^{2}-1)-x^{2}(2x)$, which collapses because $2x^{3}$ appears twice with opposite signs.
Hint 4/4
Everything cancels except $-2x$, so $f'(x)=\dfrac{-2x}{(x^{2}-1)^{2}}$.
expand only the numerator; squaring the denominator now would hide its sign, which we need later
Collect
$f'(x)=\frac{-2x}{(x^{2}-1)^{2}}$
the two cubic terms cancel, which is why this function has exactly one critical number
Answer $$f'(x)=\frac{-2x}{(x^{2}-1)^{2}}$$
Check
Sanity check at $x=2$: from the formula $f'(2)=-4/9\approx-0.44$; numerically $f(2.01)-f(1.99)$ divided by $0.02$ gives about $-0.44$ as well.
The squared denominator is always positive, so for this function the sign of $f'$ is decided by $-2x$ alone — one of the easiest sign lines you will ever build.
Notation
symbol
reads as
means
watch out
$f''(x)$
f double prime of x
the derivative of the slope function $f'$
Not $f'$ squared. It is one more differentiation, and it measures how the slope is changing.
$f'>0 \text{ on } (a,b)$
f prime is positive on the open interval from a to b
the inequality holds at every point strictly between $a$ and $b$
One test value stands for a whole piece only when $f'$ has no zero and no gap inside that piece.
$(-\infty,-1)\cup(1,\infty)$
the union of the two intervals
all $x$ below $-1$ together with all $x$ above $1$
Increasing on each of two pieces is not increasing on the union. For monotonicity answers write "on $(-\infty,-1)$ and on $(1,\infty)$".
$\lim_{x\to\infty}f(x)=L$
the limit of f of x as x tends to infinity is L
the outputs settle on the single number $L$ as $x$ runs to the right without bound
This is the horizontal asymptote statement $y=L$; the graph is allowed to cross that line at finite $x$.
$\lim_{x\to a}f(x)=\infty$
the limit is infinity
the outputs grow past every bound near $a$
Infinity is not a value: the limit does not exist, and this notation only records the way it fails.
$f(c)$
the value of f at c
the height of the graph at the critical number $c$
"The local maximum" usually means the value $f(c)$ while "where" means $c$. Report both, as the point $(c,f(c))$.
Conventions used here
Domain first
No claim about continuity, monotonicity, concavity or inflection is made at a number outside the domain of $f$. For $f(x)=1/x$ the correct sentence is "$f$ is not defined at $0$", not "$f$ is discontinuous at $0$".
Half the wrong answers in this material are statements about points that are not on the graph.
Infinite limits
$\lim f=\infty$ and "the limit does not exist" are both true when the outputs blow up; we write $\infty$ because it carries more information, and we never treat it as a number.
Writing a limit as a number that is not a number is how sign errors get laundered.
Angles
Every trigonometric function here takes radians.
The derivative formulas are false in degrees, and this section uses them inside sign lines.
Interval endpoints
Monotonicity and concavity are reported on open intervals in this section. Where $f$ is continuous at a breakpoint the closed form is also correct; pick one and stay with it inside a single answer.
Mixing the two forms inside one answer is what makes a correct sign line look like a guess.
What one derivative sign does to a whole interval
We can differentiate quickly by now; what is missing is the step from $f'(x)$, a fact about one point, to "$f$ is increasing here", a claim about every pair of points in an interval.
Say whether $f$ climbs across all of $(-2,2)$, or only at the sampled points.
Rule out a dip hiding between two samples.
Decide whether $x=-1$ is a peak, a valley, or neither.
Sample and join the dots: $f(-2)=-2$, $f(0)=0$, $f(2)=2$ climb steadily, so we draw one rising curve.
Why it fails
$f(-1)=2$ and $f(1)=-2$: between those two the graph falls four units, and the three samples never see it. More samples shrink the gaps but never prove that nothing happens inside one of them.
TheoremIncreasing/Decreasing Test
Conditions
$f$ is continuous on the interval $I$
$f$ is differentiable at every interior point of $I$
$$\boxed{\;f'>0 \text{ on } I \;\Longrightarrow\; f \text{ increasing on } I, \qquad f'\lt 0 \text{ on } I \;\Longrightarrow\; f \text{ decreasing on } I\;}$$
Positive slope at every interior point of an interval means that inside it, further right is always higher; negative slope everywhere means further right is always lower.
Where the Mean Value Theorem does the work
Take any $x_{1}\lt x_{2}$ in $I$. The Mean Value Theorem on $[x_{1},x_{2}]$ gives a $c$ between them with $f(x_{2})-f(x_{1})=f'(c)\,(x_{2}-x_{1})$. The bracket is positive, so the left side takes the sign of $f'(c)$, positive by assumption: $f(x_{2})>f(x_{1})$ for every such pair, which is what "increasing on $I$" means. One theorem, infinitely many pairs.
The chord from $A$ to $B$ has slope $4$, and the theorem promises an interior $c$ whose tangent has that same slope: one derivative value carries the whole rise from $A$ to $B$.
Looks like this, but is not
$f(x)=1/x$ has $f'(x)=-1/x^{2}\lt 0$ at every number where it is defined, so it looks like one function decreasing everywhere.
The test needs an interval, and the domain of $1/x$ is two of them. $f$ decreases on $(-\infty,0)$ and on $(0,\infty)$, yet $f(-1)=-1$ sits below $f(1)=1$: not decreasing on the union. An interval statement may never cross a gap in the domain.
x
f(x) = x³ − 3x
what the reader concludes
$-2$
$-2$
starts low
$-1$
$2$
already above the sample at $0$
$0$
$0$
the middle sample, and the one that misleads
$1$
$-2$
back below where it was at $-1$
$2$
$2$
ends high
Read rows $-2,\;0,\;2$ only and $f$ looks as if it climbs throughout. Read all five and the middle goes up four units and back down four.
Where x³ − 3x rises and where it falls
The function from the table, settled once and for all.
Given
$f(x)=x^{3}-3x$, defined for every real $x$
Find
the intervals of increase and of decrease
SolutionFind the breakpoints of the sign line
$f'(x)=3x^{2}-3=3(x-1)(x+1)$
factored, because the sign of a product is readable and the sign of a difference is not
$f'(x)=0 \iff x=-1 \text{ or } x=1$
a polynomial derivative exists everywhere, so its zeros are the only places the sign can flip
Evaluate one test value inside each piece
$f'(-2)=3(4)-3=9>0$
any number below −1 would do; −2 keeps the arithmetic small
$f'(0)=-3\lt 0$
0 is the cheapest number strictly between −1 and 1
$f'(2)=9>0$
and one above 1; three evaluations, three pieces
Translate the signs
$f' > 0 \text{ on } (-\infty,-1) \text{ and on } (1,\infty)$
positive slope throughout a piece means increasing throughout it
$f' \lt 0 \text{ on } (-1,1)$
and negative slope means decreasing — this is the drop the samples missed
Answer $$f \text{ increases on } (-\infty,-1) \text{ and } (1,\infty), \text{ decreases on } (-1,1)$$
Check
Independent check with values rather than slopes: $f(-0.5)=1.375$ and $f(0.5)=-1.375$, so moving right inside $(-1,1)$ does lower the graph, as "decreasing there" predicts.
Three evaluations of $f'$ decided the behaviour on the entire real line.
Whenever the answer is an interval, the work that earns it is always the same three moves: breakpoints, one test value per piece, translation.
Showing that x³ + x − 1 has exactly one real zero
A question about roots answered entirely by the sign of a derivative.
Given
$g(x)=x^{3}+x-1$
Find
how many real solutions $g(x)=0$ has
SolutionBound the number of roots from above
$g'(x)=3x^{2}+1$
differentiate; the point is what this expression can never be
$3x^{2}+1 \ge 1 > 0 \text{ for every } x$
a square is never negative, so the derivative never even reaches zero
$g \text{ increasing on } \mathbb{R} \Rightarrow \text{at most one zero}$
an takes each value at most once — a second root would need a return trip
Produce one root from below
$g(0)=-1\lt 0,\qquad g(1)=1>0$
we choose 0 and 1 because they are the cheapest pair with opposite signs
the Intermediate Value Theorem; continuity is what forbids jumping over the value 0
Answer $$\text{exactly one real zero, and it lies in } (0,1)$$
Check
Narrow it and the two arguments stay consistent: $g(0.6)=-0.184$ and $g(0.7)=0.043$, so the root is between $0.6$ and $0.7$; a second root anywhere would force $g$ to come back down, which needs $g'\lt 0$ somewhere, and $g'\ge 1$ everywhere.
"At most one" comes from monotonicity and "at least one" from continuity. Almost every exactly-one-root question is that pair of arguments in that order.
Checkpoint
§06.1 — how far one test value reaches●●○○○
Thirty seconds, no paper. A student computes $f'(-2)=9$ for the function below and starts writing conclusions.
Given
$f'(x)=3x^{2}-3$ for every real $x$, and $f'(-2)=9$
Find
Which claim is justified?
Hint 1/4
Ask what the piece around $x=-2$ is: how far can you walk from $-2$ before the sign of $f'$ could possibly change?
Hint 2/4
The sign of $f'$ can only change at a zero of $f'$, and $3x^{2}-3=3(x-1)(x+1)$ is zero at $-1$ and $1$.
Hint 3/4
So the piece containing $-2$ is $(-\infty,-1)$, and there $f'>0$: at $x=-2$, $f'=9$; at $x=-5$, $f'=72$.
Hint 4/4
The single test value licenses the whole piece it sits in, and stops at $-1$.
Show solutionLocate the piece
$f'=0 \iff x=\pm 1$
the only candidates for a sign change
$-2 \in (-\infty,-1)$
so this is the piece the test value belongs to
$f'>0 \text{ on } (-\infty,-1) \Rightarrow f \text{ increasing there}$
no zero of f' inside the piece, so the sign cannot flip inside it
Answer $$f \text{ increasing on } (-\infty,-1)$$
Check
Cross-check with the worked example above: the same sign line was built there from three test values, and it agrees on this piece.
A test value speaks for its own piece and stops at the next breakpoint.
⚠ Reading the direction of $f$ off the value of $f$
both live on the same sign line and the notation differs by a single prime
wrong$f(-0.5)=1.375>0 \;\Rightarrow\; f \text{ increasing at } -0.5$
right$f'(-0.5)=-2.25\lt 0 \;\Rightarrow\; f \text{ decreasing at } -0.5$
⚠ Merging two intervals across a gap in the domain
the union symbol is how the domain was written, so it gets copied into the answer
wrong$1/x \text{ decreasing on } (-\infty,0)\cup(0,\infty)$
right$1/x \text{ decreasing on } (-\infty,0) \text{ and on } (0,\infty)$
Reading a peak or a valley off the sign change
The sign line already knows where the peaks are: a piece where $f$ rises followed by a piece where it falls has a top between them.
TheoremFirst Derivative Test
Conditions
$c$ is a critical number of $f$
$f$ is continuous at $c$
$f$ is differentiable on an open interval around $c$, except possibly at $c$ itself
$$\boxed{\begin{aligned} f' : +\to- \ \text{at } c &\;\Longrightarrow\; \text{local maximum at } c\\ f' : -\to+ \ \text{at } c &\;\Longrightarrow\; \text{local minimum at } c\\ \text{no sign change} &\;\Longrightarrow\; \text{neither} \end{aligned}}$$
Climb then fall and you were on a summit; fall then climb and you were at the bottom of a valley; keep going the same way and the flat moment was only a pause.
The same curve with the sign of $f'$ drawn underneath: the graph turns over exactly where the strip changes sign, at $(-1,2)$ and at $(1,-2)$.
Looks like this, but is not
$f(x)=x^{3}$ at $c=0$: the derivative is zero, the tangent is horizontal, and the graph flattens exactly the way it does just before a peak.
$f'(x)=3x^{2}$ is positive on both sides of $0$, so nothing changes sign: the curve flattens and carries on climbing, $f(-1)=-1$ below $f(0)=0$ below $f(1)=1$. A horizontal tangent is a candidate, never a verdict.
Classifying the critical numbers of x⁴ − 4x³
Two critical numbers, and only one of them is an extremum.
Given
$f(x)=x^{4}-4x^{3}$
Find
every critical number, classified, with its value
SolutionBreakpoints
$f'(x)=4x^{3}-12x^{2}=4x^{2}(x-3)$
factoring exposes that one factor is a square, which will matter in a moment
$f'(x)=0 \iff x=0 \text{ or } x=3$
both are in the domain, so both are critical numbers
Sign line
$f'(-1)=4(1)(-4)=-16\lt 0$
a test value below 0
$f'(1)=4(1)(-2)=-8\lt 0$
and one between 0 and 3 — same sign, because 4x² cannot change sign at 0
$f'(4)=4(16)(1)=64>0$
and one above 3
Read the verdicts
$x=0:\ - \to - \Rightarrow \text{neither}$
no sign change, so the graph merely pauses on its way down
$x=3:\ - \to + \Rightarrow \text{local minimum}$
falling then rising
$f(3)=81-108=-27$
the question asks for the value as well as the place
Answer $$\text{local minimum } -27 \text{ at } x=3;\quad x=0 \text{ is a critical number but not an extremum}$$
Check
Independent check with heights: $f(-1)=5$, $f(1)=-3$, $f(2)=-16$, $f(4)=0$. The values fall through $x=0$ without bouncing and climb again after $3$, exactly as the sign line says.
One factorisation and three evaluations of $f'$ classified both critical numbers.
An even power in a factor never flips the sign. That is the whole reason $x=0$ survives as a critical number but fails as an extremum.
Checkpoint
§06.2 — an odd power still flips the sign●●○○○
Thirty seconds. The derivative below is given to you already factored, and $x=2$ is its only critical number.
Given
$f'(x)=(x-2)^{3}$ for every real $x$
Find
Classify $x=2$.
Hint 1/4
You do not need $f$ at all. The only question is what the sign of $(x-2)^{3}$ does as $x$ passes $2$.
Hint 2/4
First Derivative Test: plus to minus is a maximum, minus to plus a minimum, no change means neither.
Hint 3/4
Test values, one on each side: at $x=1$, $(1-2)^{3}=-1$; at $x=3$, $(3-2)^{3}=1$.
Hint 4/4
Minus then plus, so $x=2$ is a local minimum.
Show solutionSign on each side
$f'(1)=(-1)^{3}=-1\lt 0$
odd powers keep the sign of the base
$f'(3)=(1)^{3}=1>0$
and on the other side the base is positive
$-\to+ \Rightarrow \text{local minimum at } x=2$
the First Derivative Test read off the two signs
Answer $$\text{local minimum at } x=2$$
Check
Contrast with an even power: $(x-2)^{2}$ gives $+1$ on both sides, so that one would be "neither" — the parity of the exponent is what decides.
Odd exponent flips the sign, even exponent does not; that single fact answers most sign-line questions with repeated factors.
⚠ Treating an even power as a sign change
every factor written down looks like a breakpoint, and $x^{2}=0$ at $x=0$ does mark one
wrong$f'(x)=x^{2}(x-3):\ \text{sign changes at } x=0 \text{ and } x=3$
right$x^{2}\ge 0 \text{ on both sides} \Rightarrow \text{only } x=3 \text{ flips}$
⚠ Reporting the location when the value was asked for
the sign line ends with a number on the axis, and that number is the location
wrong$\text{the local minimum is } 3$
right$\text{the local minimum value is } f(3)=-27, \text{ at } x=3$
Bending: what the sign of the second derivative adds
Knowing that a graph rises says nothing about how it rises: two curves can climb between the same two points, one bulging upward and one sagging.
TheoremConcavity Test, and what an inflection point is
Conditions
$f''$ exists on the interval $I$
for an inflection point: $f$ is continuous at the point and the concavity changes there
$$\boxed{\;f''>0 \text{ on } I \Rightarrow f \text{ concave up on } I, \qquad f''\lt 0 \text{ on } I \Rightarrow f \text{ concave down on } I\;}$$
$f''$ is the derivative of the slope, so $f''>0$ means the slopes are getting larger as you move right: the curve keeps turning to the left and holds water, staying above each of its tangent lines. An inflection point is a point of the graph where that bending changes side.
The same curve again, now cut by the sign of $f''$: it sags to the left of the origin, bulges to the right, and at $(0,0)$ the tangent line crosses instead of staying on one side.
Looks like this, but is not
$f(x)=x^{4}$ at $c=0$: here $f''(0)=0$, the graph flattens, and it looks exactly like the moment where bending switches.
$f''(x)=12x^{2}$ is positive on both sides of $0$, so the curve is throughout and never changes side. A zero of $f''$ is a candidate inflection point in precisely the way a zero of $f'$ was a candidate extremum: it has to be confirmed by a sign change.
Concavity and inflection points of x⁴ − 4x³
The same function as the previous block, one derivative further up.
Given
$f(x)=x^{4}-4x^{3}$, so $f'(x)=4x^{2}(x-3)$
Find
the intervals of concavity and every inflection point
SolutionBreakpoints of the second sign line
$f''(x)=12x^{2}-24x=12x(x-2)$
differentiate f′ and factor at once; the factors are both to the first power, so both can flip a sign
$f''(x)=0 \iff x=0 \text{ or } x=2$
the two candidates
Sign line for f''
$f''(-1)=12(-1)(-3)=36>0$
below 0
$f''(1)=12(1)(-1)=-12\lt 0$
between 0 and 2
$f''(3)=12(3)(1)=36>0$
above 2
Translate, then confirm the inflection points
$\text{concave up on } (-\infty,0) \text{ and } (2,\infty), \text{ down on } (0,2)$
positive bends up, negative bends down
$(0,0) \text{ and } (2,-16) \text{ are inflection points}$
the sign of f'' changes at both, and f is continuous there, so both survive
$f(0)=0,\qquad f(2)=16-32=-16$
an inflection point is a point of the graph, so it needs its height
Answer $$\text{concave up on } (-\infty,0) \text{ and } (2,\infty); \text{ concave down on } (0,2); \text{ inflection points } (0,0) \text{ and } (2,-16)$$
Check
Independent check from the definition rather than from $f''$: the slopes themselves are $f'(1)=-8$, $f'(2)=-16$, $f'(2.5)=-12.5$. They decrease from $1$ to $2$ and increase after $2$ — decreasing slope is exactly , so the switch at $x=2$ is real.
$x=0$ is the number the First Derivative Test refused to classify, and here is what it is instead: an inflection point that happens to have a horizontal tangent. The curve pauses, keeps descending, and changes the way it bends.
Checkpoint
§06.3 — which zeros of the second derivative survive●●●○○
Thirty seconds. $f$ is twice differentiable on the whole real line and its second derivative is handed to you factored.
Given
$f''(x)=x^{2}(x-1)$ for every real $x$
Find
Where does the graph of $f$ have an inflection point?
Hint 1/4
Two candidates are visible. The question is which of them the second derivative actually crosses zero at, rather than merely touching.
Hint 2/4
An inflection point needs a change of sign of $f''$, not just a zero of $f''$.
Hint 3/4
Test values: $f''(-1)=(1)(-2)=-2$, $f''(0.5)=(0.25)(-0.5)=-0.125$, $f''(2)=(4)(1)=4$.
Hint 4/4
The sign is negative on both sides of $0$ and flips only at $1$, so $x=1$ is the only inflection point.
Show solutionTest each side of each candidate
$f''(-1)=-2\lt 0,\quad f''(0.5)=-0.125\lt 0$
same sign on both sides of 0, because a square cannot be negative
$f''(2)=4>0$
and the sign does flip across 1
$\text{inflection point at } x=1 \text{ only}$
a sign change plus continuity is the whole definition
Answer $$\text{inflection point at } x=1 \text{ only}$$
Check
Same structure as the extremum question one block earlier, one derivative higher: even exponents never flip a sign, and that is what disqualifies $x=0$.
Zeros of $f''$ are candidates. Only the sign changes are inflection points.
⚠ Calling every zero of $f''$ an inflection point
it mirrors the equally wrong habit of calling every zero of $f'$ an extremum
wrong$f''(0)=0 \ \Rightarrow\ (0,f(0)) \text{ is an inflection point}$
right$f(x)=x^{4}: f''=12x^{2}>0 \text{ on both sides} \Rightarrow \text{no inflection}$
⚠ Announcing an inflection point where the function is not defined
the sign of $f''$ really does flip across a vertical asymptote, and the sign line shows it
wrong$f(x)=\frac{x^{2}}{x^{2}-1}: \text{ inflection point at } x=1$
right$1 \notin \operatorname{dom} f \Rightarrow \text{ no point of the graph there to inflect}$
The one line shortcut, and when it is silent
If $f''$ is already on the page for the concavity question, it can often classify a critical number without a sign line at all.
TheoremSecond Derivative Test
Conditions
$f'(c)=0$
$f''$ is continuous on an interval around $c$
$$\boxed{\;f''(c)>0 \Rightarrow \text{local minimum at } c, \qquad f''(c)\lt 0 \Rightarrow \text{local maximum at } c, \qquad f''(c)=0 \Rightarrow \text{no conclusion}\;}$$
At a horizontal tangent a curve that bends upward can only be sitting in a valley, and one that bends downward can only be sitting on a hill. If it does not bend at all there, the test has nothing to say and you go back to the sign of $f'$.
Three functions with the same two facts at the origin — horizontal tangent and zero second derivative — and three different verdicts. This is why the third line of the test must say nothing at all.
Looks like this, but is not
The third line of the theorem looks like a third verdict: "$f''(c)=0$, therefore neither a maximum nor a minimum".
It is not a verdict, it is a gap. $x^{4}$ has a minimum at $0$, $-x^{4}$ has a maximum there, $x^{3}$ has neither, and all three satisfy $f'(0)=f''(0)=0$. The test does not say "neither"; it says nothing.
Classifying the critical numbers of x³ − 3x in two evaluations
The function from the first block, done the short way.
Given
$f(x)=x^{3}-3x$, so $f'(x)=3x^{2}-3$ and the critical numbers are $\pm 1$
Find
the classification of each critical number
SolutionGet the second derivative once
$f''(x)=6x$
one differentiation serves both critical numbers
Evaluate it at each critical number
$f''(-1)=-6\lt 0 \Rightarrow \text{local maximum at } -1$
bending down at a horizontal tangent can only be a hilltop
$f''(1)=6>0 \Rightarrow \text{local minimum at } 1$
bending up at a horizontal tangent can only be a valley floor
$f(-1)=2,\qquad f(1)=-2$
values, because "where" and "how high" are different questions
Answer $$\text{local maximum } 2 \text{ at } x=-1;\quad \text{local minimum } -2 \text{ at } x=1$$
Check
Independent check against the first block: the sign line there gave a peak at $-1$ and a valley at $1$ from three evaluations of $f'$. Same verdict, different fact, and the heights $2$ and $-2$ match the figure.
Two evaluations instead of a three-piece sign line — but only because $f''$ was one line to write.
Use whichever is cheaper. If $f''$ is a mess, or it vanishes at the critical number, the sign line of $f'$ never fails.
Checkpoint
§06.4 — the silent case, on a function you already know●●○○○
Thirty seconds, on a function met earlier in this section: its critical number at $x=0$ turned out not to be an extremum.
Given
$f(x)=x^{4}-4x^{3}$, with $f'(0)=0$ and $f''(0)=0$
Find
What does the Second Derivative Test conclude about $x=0$?
Hint 1/4
Match the two given numbers against the three lines of the theorem before doing anything else.
Hint 2/4
The three lines are $f''(c)>0$, $f''(c)\lt 0$, and $f''(c)=0$; only the first two carry a verdict.
Hint 3/4
Here $f''(0)=0$ exactly, which is the third line: the test returns nothing and the sign of $f'$ has to decide.
Hint 4/4
Nothing follows from this test; the sign line of $f'$ showed no change at $0$, so $x=0$ is neither.
Show solutionMatch the case
$f''(0)=0 \Rightarrow \text{test gives no conclusion}$
the third line of the theorem is an absence of information, not a verdict
$f'(x)=4x^{2}(x-3)\lt 0 \text{ on both sides of } 0$
so the fallback test does answer it: no sign change
$\Rightarrow x=0 \text{ is neither a maximum nor a minimum}$
the verdict comes from the First Derivative Test, not from f''
Answer $$\text{the test concludes nothing; } f' \text{ shows } x=0 \text{ is neither}$$
Check
The three curves in the figure above have the same two numbers at the origin and three different answers, which is exactly why no verdict can be read from $f''(c)=0$.
Silence is a fourth outcome of the test, and it costs you one sign line.
⚠ Reading $f''(c)=0$ as "neither"
the third line of the theorem sits next to two verdicts, so it looks like a third verdict
wrong$f''(c)=0 \Rightarrow \text{no extremum at } c$
right$f''(c)=0 \Rightarrow \text{test silent; check the sign change of } f'$
⚠ Using the test at a number that is not critical
$f''(c)\lt 0$ feels like bad news for a maximum wherever it happens
wrong$f''(2)\lt 0 \Rightarrow \text{local maximum at } 2$
right$\text{need } f'(2)=0 \text{ first; alone, } f''(2)\lt 0 \text{ only means concave down there}$
Far out: limits at infinity and horizontal asymptotes
Sign lines describe the middle of the picture. The two ends need a different question: what happens to $f(x)$ when $x$ runs away?
DefinitionLimits at infinity and horizontal asymptotes
Conditions
$f$ is defined on an interval of the form $(a,\infty)$ for the right-hand statement, $(-\infty,a)$ for the left-hand one
$$\boxed{\;\lim_{x\to\infty}\frac{1}{x^{r}}=0 \ (r>0), \qquad \lim_{x\to\pm\infty}f(x)=L \;\Longrightarrow\; y=L \text{ is a horizontal asymptote}\;}$$
If the outputs settle on a single number as $x$ runs off to the right, or to the left, then the level line at that height is the graph's long-run guide rail. There are at most two, one per direction, and the graph is allowed to cross them at finite $x$.
The method that follows from it
Divide the numerator and the denominator by the highest power of $x$ that appears in the denominator. Every term then becomes a constant or a power of $1/x$, and each power of $1/x$ goes to $0$. Nothing else is needed for the expressions in this section.
The outputs pile up under the line $y=1$ from both directions and never reach it: at $x=2$ the graph is already at $0.8$, at $x=10$ at $0.99$.
Looks like this, but is not
$f(x)=x+\dfrac{1}{x}$ has $1/x\to 0$ as $x\to\infty$, so it looks as if the graph settles down.
Only the second term settles; the first runs off to infinity, so there is no horizontal asymptote. What the graph does hug is the slanted line $y=x$, and a line like that is a — the usual sign of one is a numerator whose degree is exactly one above the denominator's.
x
f(x)
gap 1 − f(x) = 1/(x² + 1)
$1$
$0.5$
$0.5$
$2$
$0.8$
$0.2$
$5$
$0.961538$
$0.038462$
$10$
$0.990099$
$0.009901$
$100$
$0.99990001$
$0.00009999$
The gap is exactly $1/(x^{2}+1)$, so multiplying $x$ by ten divides the gap by about a hundred. "Approaches $1$" is not vague: it is this column going to zero.
Horizontal asymptotes of (3x² − x)/(2x² + 5)
The standard rational case, done by dividing rather than by guessing.
Given
$f(x)=\dfrac{3x^{2}-x}{2x^{2}+5}$
Find
$\lim_{x\to\infty}f(x)$, $\lim_{x\to-\infty}f(x)$, and the horizontal asymptotes
SolutionDivide by the highest power of the denominator
quotient of limits, legal because the denominator limit 2 is not zero
$\lim_{x\to-\infty}f(x)=\frac{3}{2}$
the same computation: 1/x and 5/x² go to 0 from the other side, and even powers do not care
Answer $$y=\tfrac{3}{2} \text{ is the only horizontal asymptote, in both directions}$$
Check
Independent check by plugging in a large number: at $x=1000$ the quotient is $2\,999\,000/2\,000\,005 \approx 1.4995$, which is $3/2$ to three decimals.
Equal degrees give the ratio of the leading coefficients. You can quote that shortcut once you have seen why it is true, but write the division on the exam paper, because the shortcut is what fails on the next example.
Two different asymptotes: √(4x² + 1)/(x + 2)
A root in the numerator, and the two directions stop agreeing.
for large negative x the numerator carries the extra minus sign
$\longrightarrow -2$
same arithmetic, opposite sign
Answer $$y=2 \text{ as } x\to\infty,\qquad y=-2 \text{ as } x\to-\infty$$
Check
Independent check with a number: at $x=-1000$ the numerator is $\sqrt{4\,000\,001}\approx 2000.0003$ and the denominator is $-998$, giving $-2.004$ — near $-2$, not $+2$.
One absolute value, two directions, two different asymptotes.
Any time a square root of an even power meets a negative direction, write $\sqrt{x^{2}}=\vert x\vert$ explicitly. It is the single most common lost mark in this material.
Checkpoint
§06.5 — the sign that hides inside a square root●●●○○
Thirty seconds. The direction is the negative one, which is the whole point of the question.
cancel x, keeping the minus that the absolute value produced
$\longrightarrow -3$
2/x² goes to 0 and the square root of 9 is 3
Answer $$-3$$
Check
Sign check without any algebra: the numerator is a square root, hence positive, and the denominator is large and negative, so the quotient must be negative.
In a limit at minus infinity, check the sign first and the size second.
⚠ Writing √(x²) = x in the negative direction
the identity is true for the positive half of the axis, which is where it was learned
wrong$\sqrt{x^{2}}=x \text{ for all } x$
right$\sqrt{x^{2}}=\vert x\vert, \text{ so } \sqrt{x^{2}}=-x \text{ when } x\lt 0$
⚠ Treating a quotient of two growing quantities as 1
both parts run to infinity, so they feel like they should cancel
Every piece of the picture is now available; what is missing is an order to collect them in, so that nothing is forgotten at the point where the clock is running.
MethodThe order of a curve sketch
Conditions
$f$ is given by a formula, not by a graph
each step uses only what the steps before it produced
Find where the function lives, where it meets the axes, whether one half mirrors the other, what it does at the edges of its domain and far away, then which way it goes and which way it bends — and only then draw. The full step list, with what each step is for, is in the checklist box below this block.
The finished picture of the worked example: three branches cut apart by the vertical asymptotes, the two outer ones settling toward $y=1$ while the middle one falls to $-\infty$ at both ends, one local maximum at the origin — and no inflection point anywhere, even though the bending flips twice.
Looks like this, but is not
The three branches of $x^{2}/(x^{2}-1)$ look like three separate graphs, so it is tempting to analyse the middle one on its own.
It is one function with one formula. What splits it is the domain, not the rule: a single computation of $f'$ and a single computation of $f''$ govern all three branches at once, and the branches differ only in which piece of the sign line they sit on.
A full sketch of x² / (x² − 1)
The checklist, run from the top, on one rational function.
Given
$f(x)=\dfrac{x^{2}}{x^{2}-1}$
$f'(x)=\dfrac{-2x}{(x^{2}-1)^{2}}$ and $f''(x)=\dfrac{6x^{2}+2}{(x^{2}-1)^{3}}$
Find
domain, intercepts, symmetry, asymptotes, monotonicity, concavity, and the sketch
Independent check of the strangest claim, the maximum at height $0$: $f(\pm 0.5)=-\tfrac{1}{3}$ and $f(0)=0$, and $0>-\tfrac{1}{3}$, so the origin really is the high point of its neighbourhood even though the graph is at zero there.
Two derivatives, four limits, three test signs — and the picture is forced, not guessed.
A rational sketch is decided by three sign facts and four limits. The checklist exists so that no step needs something a later step has not produced yet.
Checkpoint
§06.6 — concavity change without an inflection point●●●○○
Thirty seconds, on the function that was just sketched. A classmate writes the sentence below in an exam answer.
Given
$f(x)=\dfrac{x^{2}}{x^{2}-1}$, concave down on $(-1,1)$ and concave up outside $[-1,1]$
Find
True or false: "$f$ therefore has no inflection point at all".
Hint 1/4
Check the definition of an inflection point word by word, including the part about where the point has to be.
Hint 2/4
An inflection point is a point of the graph: the number must be in the domain and $f$ must be continuous there, and then the concavity must change.
Hint 3/4
The sign of $f''$ changes only across $x=-1$ and $x=1$, and neither number is in the domain, since $f(\pm 1)$ would need division by $1-1=0$.
Hint 4/4
No number is left that satisfies the whole definition, so the sentence is true: the graph has no inflection point anywhere.
Show solutionTest the definition at the only candidates
$f''(x)=\dfrac{6x^{2}+2}{(x^{2}-1)^{3}}$
the numerator is never zero, so the sign of $f''$ can only change where $x^{2}-1$ does
$f(\pm 1)\ \text{undefined, since } (\pm 1)^{2}-1=0$
the first requirement of the definition fails before concavity is even considered
$\Rightarrow \text{no point } (\pm 1, f(\pm 1)) \text{ exists}$
an inflection point is a point of the graph, not a place on the x axis
$\text{claim is true}$
the concavity changes twice, but both times across a gap, so no number qualifies
Answer $$\text{true}$$
Check
Compare with a genuine inflection point, $(2,-16)$ for $x^{4}-4x^{3}$: there the function is defined, continuous, and the bending changes — all three, not two out of three.
Two conditions, in this order: is the number in the domain, and does the concavity change. Skip the first and the sign line will happily hand you two inflection points that do not exist.
⚠ Letting an interval statement cross a vertical asymptote
the sign of $f'$ genuinely is the same on both sides, so the two pieces look joinable
wrong$f'>0 \text{ for } x\lt 0 \Rightarrow f \text{ increasing on } (-\infty,0)$
right$f \text{ increasing on } (-\infty,-1) \text{ and on } (-1,0): f(-2)=\tfrac{4}{3} > f(-0.5)=-\tfrac{1}{3}$
⚠ Treating a horizontal asymptote as a wall
the vertical ones really are walls, and the word asymptote is the same
wrong$\text{the graph can never meet } y=L$
right$\text{it may cross } y=L \text{ at finite } x; \text{ only the far ends are controlled}$
Building a sign line
Any time the answer is an interval: increasing, decreasing, concave up, concave down.
Domain
Mark every number missing from the domain of $f$. These cut the line even though they are not critical numbers.
Breakpoints
Solve $g(x)=0$ for the derivative $g$ you are studying ($f'$ or $f''$), and add the points where $g$ is undefined.
Factor
Write $g$ as a product or quotient of factors. An unfactored polynomial has no readable sign.
One test value per piece
Pick a convenient number strictly inside each piece and evaluate. Do not read the sign off the shape of the formula.
Translate
For $f'$: $+$ is rising, $-$ is falling. For $f''$: $+$ bends up, $-$ bends down.
Report
One interval per piece, joined by "and", never by a union symbol, and never across a gap in the domain.
Where it goes wrong
Forgetting a domain gap, which merges two pieces that must stay apart.
Testing at a breakpoint instead of strictly inside a piece, which gives $0$ and decides nothing.
Guessing the sign of a factor like $x^{2}$, which never changes sign, from its position in the product.
Choosing between the two tests
You have a critical number $c$ and need a verdict: maximum, minimum or neither.
Is $f''$ cheap?
If $f''$ is easy to write down, evaluate $f''(c)$ first: one number often ends the question.
Read the second derivative
$f''(c)>0$ gives a local minimum, $f''(c)\lt 0$ a local maximum.
If $f''(c)=0$, stop using it
The test is silent, not negative. Go back to the sign of $f'$ on both sides of $c$.
If $f'(c)$ does not exist
The second derivative test cannot even start; only the sign change of $f'$ can classify such a point.
Report the point
Give the location $c$ and the value $f(c)$, because the two are different answers to different questions.
Where it goes wrong
Using $f''(c)$ at a number where $f'(c)\ne 0$: bending alone classifies nothing.
Reading the silent case $f''(c)=0$ as "neither maximum nor minimum".
Reporting $c$ when the question asked for the maximum value $f(c)$.
The sketching checklist
"Analyse and sketch the graph of ..." — the standard long question of this material.
Domain
Where is $f$ defined? Every excluded number is a candidate vertical asymptote.
Intercepts
Set $x=0$ for the $y$ intercept, solve $f(x)=0$ for the $x$ intercepts, if that is cheap.
Symmetry
$f(-x)=f(x)$ halves the work (mirror in the $y$ axis); $f(-x)=-f(x)$ halves it too (half turn about the origin).
Asymptotes
One-sided limits at the domain gaps give the vertical ones; $x\to\pm\infty$ gives the horizontal ones.
Sign line for $f'$
Intervals of increase and decrease, then classify each critical number.
Sign line for $f''$
Intervals of concavity, then keep only the sign changes that happen at points of the graph.
A few exact points
Plot the intercepts, the extrema, the inflection points and one point on each far branch.
Draw branch by branch
Each piece of the domain is drawn separately; a curve never crosses a vertical asymptote.
Where it goes wrong
Drawing first and computing afterwards, which turns the sign lines into decoration.
Joining two branches across a vertical asymptote into one continuous curve.
Stopping at the intervals and never producing the picture the question asked for.
Scaffolding comes off
The common skeleton
Write down the domain, and mark every number missing from it.
Differentiate and factor; solve $f'(x)=0$ and note where $f'$ fails to exist.
Keep as critical numbers only the breakpoints that are in the domain.
Evaluate $f'$ at one convenient number strictly inside each piece.
Translate the signs into rising and falling, then classify each critical number and give the value $f(c)$.
1 · fully worked
Fully worked: analyse f(x) = x³ − 6x² + 9x + 1
Every line of the skeleton written out, with its reason.
Given
$f(x)=x^{3}-6x^{2}+9x+1$
Find
intervals of increase and decrease, and every local extremum with its value
SolutionDomain and breakpoints
$\operatorname{dom} f=\mathbb{R}$
a polynomial: nothing is missing, so the sign line has no gaps
$f'(x)=3x^{2}-12x+9=3(x-1)(x-3)$
differentiate and factor in one move; the factored form is the one that can be read
$f'(x)=0 \iff x=1 \text{ or } x=3$
both are in the domain, so both are critical numbers
One test value per piece
$f'(0)=3(-1)(-3)=9>0$
0 is the cheapest number below 1
$f'(2)=3(1)(-1)=-3\lt 0$
2 sits strictly between the two breakpoints
$f'(4)=3(3)(1)=9>0$
and 4 is above 3
Translate and classify
$\nearrow \text{ on } (-\infty,1),\ \searrow \text{ on } (1,3),\ \nearrow \text{ on } (3,\infty)$
Answer $$\text{max } 5 \text{ at } x=1;\quad \text{min } 1 \text{ at } x=3$$
Check
Independent check with heights only: $f(0)=1$, $f(1)=5$, $f(2)=3$, $f(3)=1$, $f(4)=5$ — up, down, down, up, exactly the shape the sign line predicted.
Notice that the local minimum value 1 equals $f(0)$: local says nothing about the rest of the line.
2 · you write the reasoning
Easier on purpose — the numbers are smaller and the factoring is immediate. The steps are already written; your job is the reason column. Say out loud why each line is allowed before opening the one underneath it. Analyse $f(x)=x^{3}-3x^{2}$.
$f'(x)=3x^{2}-6x=3x(x-2)$
reasoning
Factoring is not decoration: $3x^{2}-6x$ has no readable sign, while $3x(x-2)$ has one on each piece of the line.
$f'(x)=0 \iff x=0 \text{ or } x=2$
reasoning
A polynomial derivative exists everywhere, so the zeros of $f'$ are the only candidates; both are in the domain, so both count.
$f'(-1)=9>0,\quad f'(1)=-3\lt 0,\quad f'(3)=9>0$
reasoning
One value per piece is enough because $f'$ is continuous and has no zero inside a piece, so it cannot change sign there without passing through zero.
local maximum $f(0)=0$ at $x=0$; local minimum $f(2)=-4$ at $x=2$
reasoning
Plus to minus at $0$ is a summit and minus to plus at $2$ is a valley; $f(2)=8-12=-4$ supplies the height that the sign line alone does not give.
3 · find the buried error
Harder, and this time the work is done for you — badly. Exactly two of the four steps below are wrong. Find both. The problem: find and classify the local extrema of $f(x)=\dfrac{x}{(x-2)^{2}}$.
Step 1. $f'(x)=\dfrac{(x-2)^{2}-x\cdot 2(x-2)}{(x-2)^{4}}=\dfrac{-x-2}{(x-2)^{3}}$ — quotient rule, then cancel one factor of $(x-2)$.
Step 2. $f'=0$ at $x=-2$, and $f'$ is undefined at $x=2$, so the critical numbers are $x=-2$ and $x=2$.
Step 3. $f'(-3)=-\tfrac{1}{125}\lt 0$, $f'(0)=\tfrac{1}{4}>0$, $f'(3)=-5\lt 0$ — one test value per piece.
Step 4. At $x=-2$ the sign goes $-\to+$, so there is a local maximum, $f(-2)=-\tfrac{1}{8}$; at $x=2$ it goes $+\to-$, so there is a local minimum.
the two buried errors (2)
⚠ step 2
$x=2$ is counted as a critical number. It is not: $f$ itself is undefined at $2$, so $2$ is not in the domain and cannot be a critical number — it is a vertical asymptote, and it cuts the sign line without ever being a candidate.
The definition is usually remembered as "zero or undefined", and the words "in the domain of $f$" drop off. The line of $f'$ really does have a breakpoint there, which makes it look like a candidate.
right
Critical numbers: $x=-2$ only. The number $2$ still splits the sign line, but as a hole in the domain, not as a candidate extremum.
⚠ step 4
The First Derivative Test is applied backwards at $x=-2$: the sign goes $-\to+$ there, which is a local minimum, value $f(-2)=-\tfrac{1}{8}$.
"Minus then plus" reads like something getting worse, and under time pressure the word maximum attaches to the first sign change on the page.
right
Falling before $-2$ and rising after it means the graph bottoms out there: local minimum, $f(-2)=\dfrac{-2}{(-4)^{2}}=-\dfrac{1}{8}$.
4 · the bare problem
§06.2 — the same skeleton with no scaffolding●●●○○
No steps, no hints inside the statement. The skeleton at the top of this ladder is the whole method; run it.
Given
$f(x)=\dfrac{x}{x^{2}+1}$, defined for every real $x$
Find
Find the intervals where $f$ increases and decreases, and classify every local extremum with its value.
Hint 1/4
The plan is fixed: domain, then $f'$, then breakpoints, then one test value per piece, then the verdicts.
Hint 2/4
Quotient rule for $f'$, then the First Derivative Test on each critical number.
Hint 3/4
With $u=x$ and $v=x^{2}+1$: $f'=\dfrac{1\cdot(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}$, and the denominator is a square, so only $1-x^{2}$ decides the sign.
Hint 4/4
Decreasing, increasing, decreasing across $x=-1$ and $x=1$: minimum $-\tfrac{1}{2}$ at $x=-1$ and maximum $\tfrac{1}{2}$ at $x=1$.
Show solutionDerivative and breakpoints
$\operatorname{dom} f=\mathbb{R}$
x² + 1 is never zero, so nothing is missing and the line has no gaps
Answer $$\text{min } -\tfrac{1}{2} \text{ at } x=-1;\quad \text{max } \tfrac{1}{2} \text{ at } x=1$$
Check
Independent check by symmetry: $f(-x)=-f(x)$, so whatever happens at $x=1$ must happen upside down at $x=-1$ — a maximum of $\tfrac{1}{2}$ against a minimum of $-\tfrac{1}{2}$, which is what came out.
A squared denominator is the friendliest thing in a sign line: it can never flip anything, so the numerator does all the work.
Full exam-style question
Full analysis of f(x) = 3x⁵ − 5x³exam format
The long question of this material, in the form it is usually asked: one function, four parts, and the picture at the end. Work it on paper before opening the solution.
Given
$f(x)=3x^{5}-5x^{3}$, defined for every real $x$
$f'(x)=15x^{4}-15x^{2}$
$f''(x)=60x^{3}-30x$
Find
(a) intervals of increase and decrease; (b) every local extremum with its value; (c) intervals of concavity and every inflection point; (d) the shape of the graph.
Solution(a) Sign line for f'
$f'(x)=15x^{2}(x^{2}-1)=15x^{2}(x-1)(x+1)$
factor fully: the square is the factor that will refuse to change sign
the x intercepts, which anchor the ends of the sketch
$\text{up to } (-1,2),\ \text{down through } (0,0) \text{ to } (1,-2),\ \text{up again}$
the sign lines leave exactly one shape available
Answer $$\nearrow(-\infty,-1),\ \searrow(-1,1),\ \nearrow(1,\infty);\ \text{max } 2 \text{ at } -1,\ \text{min } -2 \text{ at } 1;\ \text{inflections at } 0,\pm\tfrac{1}{\sqrt2}$$
Check
Independent check by symmetry: $f$ is odd, so the maximum at $x=-1$ and the minimum at $x=1$ have to be negatives of each other — $2$ and $-2$, which is what the arithmetic gave. Same for the two outer inflection heights, $1.24$ and $-1.24$.
Two sign lines, seven test values, and the sketch is forced.
Three critical numbers and only two extrema is the standard shape of this question. The one that is neither is always the factor with an even power.
Practice
A · concept 3 questions
1§06.2 — critical numbers are candidates, not verdicts●●○○○
A statement of the kind that opens a quiz. Decide true or false, and make sure you can produce a reason or a counterexample, since that is what the marks are for.
Given
Claim: "Every critical number of a differentiable function is either a local maximum or a local minimum."
Find
True or false, with justification.
Hint 1/4
Ask yourself whether you know a function with a horizontal tangent that keeps going the same way.
Hint 2/4
The First Derivative Test needs a change of sign of $f'$; a critical number without one is neither a maximum nor a minimum.
Hint 3/4
Take $f(x)=x^{3}$: $f'(x)=3x^{2}$, so $f'(0)=0$ and $0$ is a critical number, while $f'(x)>0$ for every other $x$.
Hint 4/4
One counterexample settles it: the claim is false.
Show solutionProduce the counterexample
$f'(x)=3x^{2},\quad f'(0)=0$
so 0 is a critical number
$f'(x)>0 \text{ for every } x\ne 0$
no sign change anywhere near 0
$f(-1)=-1\lt f(0)=0\lt f(1)=1$
and the values confirm it: the function climbs straight through
Answer $$\text{false}$$
Check
Independent check with heights rather than slopes: $f(-0.1)=-0.001$ and $f(0.1)=0.001$ straddle $f(0)=0$, so $0$ is neither the largest nor the smallest value nearby.
"Critical number" is the shortlist. The sign change is the interview.
2§06.1 — the converse of the increasing test●●●○○
The Increasing/Decreasing Test runs one way. This asks what is left of it when you read it backwards, which is the kind of thing a quiz likes.
Given
Claim: "If $f$ is increasing and differentiable on an interval $I$, then $f'(x)\ge 0$ at every interior point of $I$."
Find
True or false, with justification.
Hint 1/4
The test says positive slope implies increasing. Read it backwards and decide which of $f'>0$ and $f'\ge 0$ you can still defend.
Hint 2/4
An increasing differentiable function cannot have a negative derivative anywhere: every difference quotient $\dfrac{f(x_{0}+h)-f(x_{0})}{h}$ is positive, so the limit cannot be negative. Isolated zeros are still allowed.
Hint 3/4
Test the sharper version on $f(x)=x^{3}$ over $I=(-1,1)$: it is increasing there, yet $f'(0)=0$, so $f'>0$ would fail while $f'\ge 0$ holds.
Hint 4/4
So the claim as written is true, and it is true only because it says $\ge$ and not $>$.
Show solutionRule out a negative slope
$h>0:\ \dfrac{f(x_{0}+h)-f(x_{0})}{h}>0$
increasing means the numerator is positive when the step is positive
$h<0:\ \dfrac{f(x_{0}+h)-f(x_{0})}{h}>0$
now both numerator and denominator are negative, so the quotient is positive again
a limit of positive quantities cannot be negative, which is exactly the claim
Check that $\ge$ cannot be sharpened to $>$
$f(x)=x^{3}\ \text{increasing on } (-1,1),\quad f'(0)=0$
the limit is allowed to reach zero, so the strict version of the same sentence is false
Answer $$\text{true}$$
Check
Read the two directions side by side: $f'>0$ on $I$ gives increasing, and increasing gives $f'\ge 0$. The asymmetry is the whole content, and $x^{3}$ at the origin is the example that creates it.
One-way implications are exam material precisely because the reverse reading looks harmless. Here the reverse is true, but only with $\ge$; the same sentence with $>$ is false.
3§06.3 — one value of the second derivative●●●○○
A single number is handed to you and four conclusions are offered. Only one of them follows from that number alone.
Given
$f$ is twice differentiable everywhere and $f''(3)=0$.
Find
Which conclusion is justified?
Hint 1/4
A single value of $f''$ is a fact at one point. Which of the four options is also a statement about one point, rather than about an interval or about a change?
Hint 2/4
An inflection point needs the sign of $f''$ to change across the number; a maximum or minimum needs $f'(c)=0$ first. Neither is given.
Hint 3/4
Given data again: only $f''(3)=0$, with no information about $f'(3)$ and none about $f''$ anywhere else.
Hint 4/4
So nothing about shape follows yet: $3$ is a candidate inflection point and the sign of $f''$ on both sides has to decide.
Show solutionTest each candidate claim
$f(x)=x^{4}:\ f''(0)=0 \text{ and no inflection at } 0$
the standard counterexample: a zero of f″ need not be an inflection point
$\text{extremum needs } f'(3)=0$
and nothing was said about f′, so no extremum claim can be made
$\Rightarrow \text{candidate only}$
the honest conclusion is a to-do item, not a verdict
Answer $$\text{no shape conclusion; check the sign of } f'' \text{ on both sides}$$
Check
Independent check: $x^{4}$ has $f''(0)=0$ with the graph concave up on both sides, while $x^{3}$ has $f''(0)=0$ with a genuine inflection — same fact, opposite outcomes.
Every "the derivative is zero here" fact in this section is a candidate that has to be confirmed by a sign change.
B · computation 5 questions
1§06.1 — intervals and extrema of a cubic●●○○○
The bread-and-butter computation of this section. Everything is decided by one factored derivative and three test values.
Given
$f(x)=2x^{3}-9x^{2}+12x$
Find
(a) Find the intervals of increase and decrease.
(b) Classify every local extremum and give its value.
Hint 1/4
Nothing here needs $f$ itself until the very last line, when the values are asked for.
Hint 2/4
Differentiate, factor, solve $f'=0$, one test value per piece, then the First Derivative Test.
Hint 3/4
With $f(x)=2x^{3}-9x^{2}+12x$: $f'(x)=6x^{2}-18x+12=6(x-1)(x-2)$, so the breakpoints are $1$ and $2$.
Hint 4/4
Increasing outside $[1,2]$ and decreasing inside it: maximum $f(1)=5$, minimum $f(2)=4$.
Show solutionBreakpoints
$f'(x)=6x^{2}-18x+12=6(x-1)(x-2)$
factor out the 6 first; the quadratic then factors by inspection
Answer $$\text{max } 5 \text{ at } x=1;\quad \text{min } 4 \text{ at } x=2$$
Check
Independent check: the local maximum $5$ must sit above the local minimum $4$ for a curve that falls from one to the other, and $f(1.5)=6.75-20.25+18=4.5$ lies between them, as it must.
Local maximum above local minimum is a free sanity check whenever the two are joined by one falling piece.
2§06.3 — concavity and inflection points of a quartic●●●○○
Same machinery, one derivative higher. The function is even, which is worth noticing before you compute anything.
Given
$f(x)=x^{4}-6x^{2}+5$
Find
(a) Find the intervals of concavity.
(b) Give every inflection point as a point, not just an $x$ value.
Hint 1/4
The question is about bending, so the object of study is $f''$, not $f'$.
Hint 2/4
Concavity Test: $f''>0$ bends up, $f''\lt 0$ bends down; an inflection point needs a sign change of $f''$ at a number in the domain.
Hint 3/4
With $f(x)=x^{4}-6x^{2}+5$: $f'(x)=4x^{3}-12x$ and $f''(x)=12x^{2}-12=12(x-1)(x+1)$.
Hint 4/4
Concave up outside $[-1,1]$, concave down inside, and both $\pm 1$ are inflection points at height $f(\pm 1)=0$.
Show solutionSecond derivative, factored
$f'(x)=4x^{3}-12x$
power rule
$f''(x)=12x^{2}-12=12(x-1)(x+1)$
and again, then factor so the sign is readable
Sign line
$f''(-2)=12(3)(-1)\cdot(-1)=36>0$
below −1, where both factors are negative
$f''(0)=-12\lt 0$
between −1 and 1
$f''(2)=36>0$
above 1
Confirm the inflection points
$\text{sign changes at } x=\pm 1 \text{ and } f \text{ is a polynomial}$
so both survive: continuity is automatic here
$f(\pm 1)=1-6+5=0$
the height each inflection point sits at
Answer $$\text{up on } (-\infty,-1) \text{ and } (1,\infty);\ \text{down on } (-1,1);\ \text{inflections } (\pm 1,0)$$
Check
Independent check from the definition of concavity: the slopes are $f'(-2)=-8$ and $f'(-1)=8$, rising before $-1$, which is concave up; then $f'(-0.5)=5.5$ and $f'(0.5)=-5.5$, falling inside $(-1,1)$, which is concave down.
Every inflection point is reported as a point $(c,f(c))$; a bare $x$ value is half an answer.
3§06.4 — second derivative test on a function with a domain gap●●●○○
A rational function whose two extrema sit in the wrong order — the local maximum is below the local minimum. That is not an error; it is what "local" means.
Given
$f(x)=x+\dfrac{4}{x}$, defined for $x\ne 0$
Find
(a) Find the critical numbers.
(b) Classify each one with the Second Derivative Test and give its value.
Hint 1/4
Rewrite the second term as a power before differentiating, and keep the domain in view.
Hint 2/4
$\frac{d}{dx}\left(4x^{-1}\right)=-4x^{-2}$; then the Second Derivative Test: $f''(c)>0$ is a minimum, $f''(c)\lt 0$ a maximum.
Hint 3/4
With $f(x)=x+4x^{-1}$: $f'(x)=1-\dfrac{4}{x^{2}}$, which is zero when $x^{2}=4$, and $f''(x)=\dfrac{8}{x^{3}}$.
Hint 4/4
At $x=2$: $f''=1>0$, minimum $f(2)=4$. At $x=-2$: $f''=-1\lt 0$, maximum $f(-2)=-4$.
Show solutionCritical numbers
$f'(x)=1-4x^{-2}=1-\frac{4}{x^{2}}$
power rule on the rewritten term
$f'(x)=0 \iff x^{2}=4 \iff x=\pm 2$
both are in the domain
$x=0 \text{ is not a critical number}$
f is undefined there, so it cannot be one; it is a vertical asymptote
Answer $$\text{min } 4 \text{ at } x=2;\quad \text{max } -4 \text{ at } x=-2$$
Check
Independent check that the strange ordering is real: $f(-1)=-5$ and $f(-4)=-5$ both sit below $f(-2)=-4$, so $-4$ is genuinely the largest value in its own neighbourhood even though $f(2)=4$ is far bigger elsewhere on the graph.
"Local" is a promise about a neighbourhood and nothing else; the vertical asymptote at $0$ is what lets the two values sit in this order.
4§06.5 — two limits at infinity, one with a root●●●○○
Two short limits. The first is the routine rational case; the second is the one that punishes a missing absolute value.
(b) Evaluate the limit in (b), and say explicitly where the sign comes from.
Hint 1/4
Neither limit is a substitution. In both, decide first which power of $x$ dominates the denominator.
Hint 2/4
Divide numerator and denominator by the highest power in the denominator; for a root, first write $\sqrt{x^{2}}=\vert x\vert$ and only then choose $x$ or $-x$.
Hint 3/4
In (a) divide by $x^{3}$: $\dfrac{4-\frac{1}{x^{2}}}{\frac{1}{x^{3}}-2}$. In (b), $x\to-\infty$ means $x\lt 0$, so $\vert x\vert=-x$ and $\sqrt{x^{2}+4x}=-x\sqrt{1+\frac{4}{x}}$.
Hint 4/4
(a) tends to $\dfrac{4}{-2}=-2$; (b) tends to $-1$.
cancel x and keep the minus the absolute value produced
$\longrightarrow -1$
4/x goes to 0 and the root of 1 is 1
Answer $$(a)\ -2 \qquad (b)\ -1$$
Check
Independent check with a number: at $x=-100$, $\sqrt{10\,000-400}=\sqrt{9600}\approx 97.98$, and dividing by $-100$ gives $-0.98$ — close to $-1$ and unmistakably negative.
Whenever the direction is $-\infty$ and a square root is present, write the absolute value line explicitly; it is the step that is graded.
5§06.2 — classifying from the behaviour of f' alone●●○○○
Here you are not given a formula at all, only a description of $f'$. Everything in this section that classifies a critical number needs nothing more than this.
Given
$f$ is continuous on $\mathbb{R}$
$f'\lt 0$ on $(-\infty,2)$; $f'(2)=0$; $f'>0$ on $(2,5)$; $f'(5)=0$; $f'>0$ on $(5,\infty)$
Find
(a) List the critical numbers of $f$.
(b) Classify each one as a local maximum, a local minimum, or neither.
(c) Say on which intervals $f$ increases and decreases.
Hint 1/4
Do not try to reconstruct $f$. The classification uses only the signs on either side of each critical number.
Hint 2/4
First Derivative Test: $-\to+$ is a minimum, $+\to-$ a maximum, and no change is neither.
Hint 3/4
The data again: negative before $2$, positive between $2$ and $5$, positive again after $5$. So $2$ has a sign change and $5$ does not.
Hint 4/4
$x=2$ is a local minimum; $x=5$ is a critical number that is neither.
Show solutionCritical numbers
$f'(2)=0,\quad f'(5)=0$
the derivative exists everywhere here, so its zeros are the whole list
Classify each
$x=2:\ -\to+ \Rightarrow \text{local minimum}$
falling then rising
$x=5:\ +\to+ \Rightarrow \text{neither}$
a horizontal tangent with no change of direction: the graph pauses and continues up
Monotonicity
$\searrow \text{ on } (-\infty,2),\quad \nearrow \text{ on } (2,\infty)$
the two positive pieces join, because f is continuous at 5 and the slope only vanishes at that single point
Answer $$\text{min at } x=2;\ x=5 \text{ neither};\ \searrow(-\infty,2),\ \nearrow(2,\infty)$$
Check
Independent check against a concrete model: $f'(x)=(x-2)(x-5)^{2}$ has exactly this sign pattern — negative below $2$, positive above it, and a zero at $5$ that flips nothing.
A critical number where $f'$ keeps its sign is a flat spot on a slope, and it will appear in the sketch as a brief straightening.
C · exam level 3 questions
1§06.3 — the standard long question, start to finish●●●●○
The full four-part analysis, on a cubic chosen so that every number stays an integer. Budget about eight minutes and write the sign lines out.
Given
$f(x)=x^{3}-3x^{2}-9x+5$
Find
(a) Find the intervals of increase and decrease.
(b) Classify every local extremum and give its value.
(c) Find the intervals of concavity and every inflection point.
(d) Describe the shape of the graph in one sentence.
Hint 1/4
Two sign lines are needed, one for $f'$ and one for $f''$, and they answer different parts of the question.
Hint 2/4
Increasing/Decreasing Test for (a), First Derivative Test for (b), Concavity Test for (c).
Hint 3/4
With $f(x)=x^{3}-3x^{2}-9x+5$: $f'(x)=3x^{2}-6x-9=3(x-3)(x+1)$ and $f''(x)=6x-6=6(x-1)$.
Hint 4/4
Up to $(-1,10)$, down to $(3,-22)$, up again, with the bend switching at $(1,-6)$.
Independent check: for a cubic the inflection point must sit exactly halfway between the two critical numbers, and $\tfrac{-1+3}{2}=1$ matches the value found from $f''$.
The halfway property of a cubic's inflection point is worth remembering purely as a check on your own arithmetic.
2§06.6 — a rational function with a horizontal asymptote●●●●○
The other standard long question: a quotient, so the two ends of the picture matter as much as the middle. The denominator never vanishes, which removes vertical asymptotes from the work.
Given
$f(x)=\dfrac{4x}{x^{2}+4}$
Find
(a) State the domain and any symmetry.
(b) Find all horizontal asymptotes.
(c) Find the intervals of increase and decrease and classify every local extremum.
(d) Describe the resulting sketch.
Hint 1/4
Run the checklist in order; do not start with the derivative.
Hint 2/4
Domain from the denominator, symmetry from $f(-x)$, horizontal asymptotes from $x\to\pm\infty$, then the quotient rule and a sign line.
Hint 3/4
With $f(x)=\dfrac{4x}{x^{2}+4}$: $x^{2}+4>0$ always, $f(-x)=-f(x)$, and dividing by $x^{2}$ gives $\dfrac{4/x}{1+4/x^{2}}\to 0$; the quotient rule gives $f'(x)=\dfrac{4(4-x^{2})}{(x^{2}+4)^{2}}$.
Hint 4/4
Domain $\mathbb{R}$, odd, $y=0$ in both directions, minimum $-1$ at $x=-2$, maximum $1$ at $x=2$.
Independent check on the asymptote crossing: solving $f(x)=0$ gives $x=0$, so the graph really does meet $y=0$ at a finite point while still approaching it at both ends — the thing a vertical asymptote could never do.
A horizontal asymptote is a statement about the far ends only. This function crosses its own asymptote in the middle of the picture.
3§06.3 — counting inflection points from a factored second derivative●●●○○
An exam favourite because it can be answered in twenty seconds by someone who knows what a sign change is, and not at all by someone who counts zeros.
Given
$f$ is twice differentiable on $\mathbb{R}$ and $f''(x)=(x-1)^{2}(x-4)$
Find
How many inflection points does the graph of $f$ have?
Hint 1/4
Do not count zeros of $f''$. Count the places where $f''$ changes sign.
Hint 2/4
A factor with an even exponent touches zero without changing sign; a factor with an odd exponent changes sign.
Hint 3/4
The data again: $(x-1)^{2}$ is a square, and $(x-4)$ is to the first power. Test values: $f''(0)=(1)(-4)=-4$, $f''(2)=(1)(-2)=-2$, $f''(5)=(16)(1)=16$.
Hint 4/4
Only one sign change, at $x=4$.
Show solutionTest around each zero
$f''(0)=(1)(-4)=-4\lt 0$
left of 1
$f''(2)=(1)(-2)=-2\lt 0$
between 1 and 4: same sign, so 1 is not an inflection point
$f''(5)=(16)(1)=16>0$
right of 4: the sign flips, so 4 is one
Report
$\text{exactly one inflection point, at } x=4$
f is twice differentiable everywhere, so continuity is not in doubt
Answer $$\text{one, at } x=4$$
Check
Independent check by parity: the exponent of $(x-1)$ is even and that of $(x-4)$ is odd, and only odd exponents flip a sign — which is the same conclusion without evaluating anything.
Counting zeros of $f''$ overcounts inflection points every time there is a repeated factor.
D · interleaved 3 questions
1mixed practice — decide the method yourself●●●●○
Deliberately not labelled. Part of the exercise is deciding which tool this needs before you start writing.
Given
The curve $x^{2}+xy+y^{2}=3$
Find
Find every point on the curve where the tangent line is horizontal.
Hint 1/4
A horizontal tangent is a statement about a derivative, but $y$ is not given as a formula in $x$ here.
Hint 2/4
Differentiate both sides with respect to $x$, treating $y$ as a function of $x$, and then set the slope to zero.
Hint 3/4
From $x^{2}+xy+y^{2}=3$: $2x+y+xy'+2yy'=0$, so $y'=-\dfrac{2x+y}{x+2y}$; horizontal means the numerator is zero while the denominator is not.
Hint 4/4
$y=-2x$ together with the curve gives $3x^{2}=3$, so the points are $(1,-2)$ and $(-1,2)$.
Show solutionDifferentiate implicitly
$2x+y+xy'+2yy'=0$
product rule on xy, chain rule on y²: every y carries a y′
$y'=-\frac{2x+y}{x+2y}$
solve for y′; we solve for y′ rather than for y because the question is about the slope
Impose the condition
$y'=0 \iff 2x+y=0 \text{ and } x+2y\ne 0$
a fraction is zero exactly when its numerator is, and its denominator is not
$y=-2x$
the condition, ready to substitute into the curve
Return to the curve
$x^{2}+x(-2x)+(-2x)^{2}=3 \Rightarrow 3x^{2}=3$
the point must be on the curve as well as satisfy the slope condition
$x=\pm 1 \Rightarrow (1,-2) \text{ and } (-1,2)$
two points
$x+2y=1-4=-3\ne 0 \text{ and } -1+4=3\ne 0$
check the denominator at both, or the tangent would be vertical rather than horizontal
Answer $$(1,-2)\ \text{ and }\ (-1,2)$$
Check
Independent check that both points are on the curve: $1+(1)(-2)+4=3$ and $1+(-1)(2)+4=3$ — both give exactly $3$.
Horizontal tangent means numerator zero; vertical tangent means denominator zero. Checking the other one is not optional.
2mixed practice — decide the method yourself●●●○○
Also unlabelled, and it looks like the standard question of this section until you read the last three characters of the first line.
Given
$f(x)=x^{3}-3x$ on the closed interval $[0,3]$
Find
Find the absolute maximum and the absolute minimum of $f$ on $[0,3]$, and say where each occurs.
Hint 1/4
Notice which word is being asked for: absolute, not local. That changes what has to be compared.
Hint 2/4
On a closed bounded interval, compare the values at the critical numbers inside the interval with the values at the two endpoints.
Hint 3/4
With $f(x)=x^{3}-3x$ on $[0,3]$: $f'(x)=3x^{2}-3=0$ at $x=\pm 1$, and only $x=1$ lies inside $[0,3]$; the candidates are therefore $f(0)$, $f(1)$ and $f(3)$.
Hint 4/4
$f(0)=0$, $f(1)=-2$, $f(3)=18$: the largest is $18$ and the smallest $-2$.
a critical number outside the interval is irrelevant to this question
$\text{candidates: } x=0,\ 1,\ 3$
interior critical numbers plus both endpoints
Compare the values
$f(0)=0,\quad f(1)=1-3=-2,\quad f(3)=27-9=18$
the Closed Interval Method compares heights, so no sign line is needed
$\max=18 \text{ at } x=3,\qquad \min=-2 \text{ at } x=1$
largest and smallest of the three
Answer $$\text{absolute max } 18 \text{ at } x=3;\quad \text{absolute min } -2 \text{ at } x=1$$
Check
Independent check against the shape: this section's first block showed $f$ decreasing on $(-1,1)$ and increasing on $(1,\infty)$, so on $[0,3]$ the graph falls to $x=1$ and climbs to the right endpoint — the extremes must be $f(1)$ and $f(3)$, which is what the comparison gave.
The absolute maximum here is at an endpoint, where no derivative test would ever have looked.
3mixed practice — decide the method yourself●●●●○
Unlabelled again. It has the shape of a limit at infinity, but the first move is one you learned long before this section.
Both terms run to infinity, so the difference is not decided by either of them alone; the expression has to be rewritten before any limit is taken.
Hint 2/4
Multiply and divide by the conjugate $\sqrt{x^{2}+3x}+x$: the difference of squares kills the root in the numerator.
Hint 3/4
$\left(\sqrt{x^{2}+3x}-x\right)\cdot\dfrac{\sqrt{x^{2}+3x}+x}{\sqrt{x^{2}+3x}+x}=\dfrac{3x}{\sqrt{x^{2}+3x}+x}$, and now divide top and bottom by $x$, which is positive here.
Hint 4/4
The quotient tends to $\dfrac{3}{\sqrt{1}+1}=\dfrac{3}{2}$.
x → ∞ means x > 0, so |x| = x and the root keeps its plus sign
$\longrightarrow \frac{3}{1+1}=\frac{3}{2}$
3/x dies inside the root
Answer $$\frac{3}{2}$$
Check
Independent check with a number: at $x=1000$, $\sqrt{1\,003\,000}\approx 1001.4989$, so the difference is about $1.4989$ — that is $3/2$ to three decimals.
An $\infty-\infty$ shape is never answered where it stands: rewrite it as a quotient first, then use the dominant power.
Mistake ledger (12 entries)
⚠ Reading the direction of $f$ off the value of $f$
both live on the same sign line and the notation differs by a single prime
wrong$f(-0.5)=1.375>0 \;\Rightarrow\; f \text{ increasing at } -0.5$
right$f'(-0.5)=-2.25\lt 0 \;\Rightarrow\; f \text{ decreasing at } -0.5$
⚠ Merging two intervals across a gap in the domain
the union symbol is how the domain was written, so it gets copied into the answer
wrong$1/x \text{ decreasing on } (-\infty,0)\cup(0,\infty)$
right$1/x \text{ decreasing on } (-\infty,0) \text{ and on } (0,\infty)$
⚠ Treating an even power as a sign change
every factor written down looks like a breakpoint, and $x^{2}=0$ at $x=0$ does mark one
wrong$f'(x)=x^{2}(x-3):\ \text{sign changes at } x=0 \text{ and } x=3$
right$x^{2}\ge 0 \text{ on both sides} \Rightarrow \text{only } x=3 \text{ flips}$
⚠ Reporting the location when the value was asked for
the sign line ends with a number on the axis, and that number is the location
wrong$\text{the local minimum is } 3$
right$\text{the local minimum value is } f(3)=-27, \text{ at } x=3$
⚠ Calling every zero of $f''$ an inflection point
it mirrors the equally wrong habit of calling every zero of $f'$ an extremum
wrong$f''(0)=0 \ \Rightarrow\ (0,f(0)) \text{ is an inflection point}$
right$f(x)=x^{4}: f''=12x^{2}>0 \text{ on both sides} \Rightarrow \text{no inflection}$
⚠ Announcing an inflection point where the function is not defined
the sign of $f''$ really does flip across a vertical asymptote, and the sign line shows it
wrong$f(x)=\frac{x^{2}}{x^{2}-1}: \text{ inflection point at } x=1$
right$1 \notin \operatorname{dom} f \Rightarrow \text{ no point of the graph there to inflect}$
⚠ Reading $f''(c)=0$ as "neither"
the third line of the theorem sits next to two verdicts, so it looks like a third verdict
wrong$f''(c)=0 \Rightarrow \text{no extremum at } c$
right$f''(c)=0 \Rightarrow \text{test silent; check the sign change of } f'$
⚠ Using the test at a number that is not critical
$f''(c)\lt 0$ feels like bad news for a maximum wherever it happens
wrong$f''(2)\lt 0 \Rightarrow \text{local maximum at } 2$
right$\text{need } f'(2)=0 \text{ first; alone, } f''(2)\lt 0 \text{ only means concave down there}$
⚠ Writing √(x²) = x in the negative direction
the identity is true for the positive half of the axis, which is where it was learned
wrong$\sqrt{x^{2}}=x \text{ for all } x$
right$\sqrt{x^{2}}=\vert x\vert, \text{ so } \sqrt{x^{2}}=-x \text{ when } x\lt 0$
⚠ Treating a quotient of two growing quantities as 1
both parts run to infinity, so they feel like they should cancel
⚠ Letting an interval statement cross a vertical asymptote
the sign of $f'$ genuinely is the same on both sides, so the two pieces look joinable
wrong$f'>0 \text{ for } x\lt 0 \Rightarrow f \text{ increasing on } (-\infty,0)$
right$f \text{ increasing on } (-\infty,-1) \text{ and on } (-1,0): f(-2)=\tfrac{4}{3} > f(-0.5)=-\tfrac{1}{3}$
⚠ Treating a horizontal asymptote as a wall
the vertical ones really are walls, and the word asymptote is the same
wrong$\text{the graph can never meet } y=L$
right$\text{it may cross } y=L \text{ at finite } x; \text{ only the far ends are controlled}$
Formula card
Increasing/Decreasing Test
$\boxed{\;f'>0 \text{ on } I \;\Longrightarrow\; f \text{ increasing on } I, \qquad f'\lt 0 \text{ on } I \;\Longrightarrow\; f \text{ decreasing on } I\;}$
$f$ is continuous on the interval $I$; $f$ is differentiable at every interior point of $I$
First Derivative Test
$\boxed{\begin{aligned} f' : +\to- \ \text{at } c &\;\Longrightarrow\; \text{local maximum at } c\\ f' : -\to+ \ \text{at } c &\;\Longrightarrow\; \text{local minimum at } c\\ \text{no sign change} &\;\Longrightarrow\; \text{neither} \end{aligned}}$
$c$ is a critical number of $f$; $f$ is continuous at $c$; $f$ is differentiable on an open interval around $c$, except possibly at $c$ itself
Concavity Test, and what an inflection point is
$\boxed{\;f''>0 \text{ on } I \Rightarrow f \text{ concave up on } I, \qquad f''\lt 0 \text{ on } I \Rightarrow f \text{ concave down on } I\;}$
$f''$ exists on the interval $I$; for an inflection point: $f$ is continuous at the point and the concavity changes there
Second Derivative Test
$\boxed{\;f''(c)>0 \Rightarrow \text{local minimum at } c, \qquad f''(c)\lt 0 \Rightarrow \text{local maximum at } c, \qquad f''(c)=0 \Rightarrow \text{no conclusion}\;}$
$f'(c)=0$; $f''$ is continuous on an interval around $c$
Limits at infinity and horizontal asymptotes
$\boxed{\;\lim_{x\to\infty}\frac{1}{x^{r}}=0 \ (r>0), \qquad \lim_{x\to\pm\infty}f(x)=L \;\Longrightarrow\; y=L \text{ is a horizontal asymptote}\;}$
$f$ is defined on an interval of the form $(a,\infty)$ for the right-hand statement, $(-\infty,a)$ for the left-hand one
always; decisive when the direction is minus infinity
Check yourself
Close the page and write, from memory: the two sign facts that decide shape, what each one says about the graph, the exact case where the second derivative test is silent, and the two conditions a number must satisfy before you may call it an inflection point.
State why $f'(2)=9>0$ is not enough to call $f$ increasing on $(0,3)$, and what would be enough.
c-monotonicity
Give a function with a critical number that is neither a maximum nor a minimum, and show why.
c-first-derivative-test
Give a zero of $f''$ that is not an inflection point, and a concavity change that is not one either.
c-concavity
Say what the second derivative test concludes when $f''(c)=0$, in one sentence.
c-second-derivative-test
Compute $\lim_{x\to-\infty}\sqrt{9x^{2}+2}\,/\,x$ without looking, and justify the sign.
c-limits-at-infinity
List the eight checklist steps in order and say what each one contributes to the picture.
c-curve-sketching
Glossary (15 terms)
increasing functionartan fonksiyon
On an interval $I$: $x_{1}\lt x_{2}$ in $I$ always gives $f(x_{1})\lt f(x_{2})$.
azalan fonksiyon
On an interval $I$: $x_{1}\lt x_{2}$ in $I$ always gives $f(x_{1})>f(x_{2})$.
critical numberkritik nokta
A number in the domain of $f$ where $f'$ is zero or fails to exist; the only candidate for a local extremum.
sign lineişaret tablosu
The number line cut at the breakpoints of a derivative, with one evaluated test value recorded in each piece.
First Derivative Testbirinci türev testi
Classifies a critical number by how the sign of $f'$ changes across it: plus to minus is a maximum, minus to plus a minimum.
concave upyukarı bükey
On an interval where $f'$ is increasing, equivalently $f''>0$; the graph stays above each of its tangent lines.
concave downaşağı bükey
On an interval where $f'$ is decreasing, equivalently $f''\lt 0$; the graph stays below each of its tangent lines.
inflection pointdönüm noktası
A point of the graph where $f$ is continuous and the concavity changes side.
Second Derivative Testikinci türev testi
At a critical number with $f'(c)=0$: $f''(c)>0$ means a local minimum, $f''(c)\lt 0$ a local maximum, $f''(c)=0$ means no verdict.
limit at infinitysonsuzda limit
The number the outputs settle on as $x$ runs beyond every bound in the positive or the negative direction.
horizontal asymptoteyatay asimptot
A line $y=L$ with $\lim_{x\to\infty}f(x)=L$ or $\lim_{x\to-\infty}f(x)=L$; at most one per direction.
slant asymptoteeğik asimptot
A line $y=mx+b$ with $m\ne 0$ that the graph approaches far out, typical when the numerator degree is one above the denominator degree.
yerel maksimum değeri
The value $f(c)$ at a point where $f(c)\ge f(x)$ for all $x$ near $c$; the location is $c$, the value is $f(c)$.
mutlak maksimum değeri
The largest value of $f$ on the whole set under discussion, which may occur at an endpoint rather than a critical number.
monoton
Increasing throughout an interval, or decreasing throughout it, with no change of direction inside.
What comes next
§07 · Curve sketching, optimization, and Newton's method
Here the shape of a given function was the answer. Next it becomes the tool: once you can read a graph from $f'$ and $f''$, the same two sign lines tell you which box has the largest volume and which rectangle the cheapest fence — and the sketching checklist becomes step one of every optimisation problem.
Sources
James Stewart, Calculus, Metric Version, Ninth Edition — sections 3.3, 3.4 and 3.5 The tests are stated as this book states them; the worked functions here are different ones.
Course syllabus, week 6: Applications of Differentiation 3.3, 3.4, 3.5 The assessment weights quoted on the card come from the same syllabus.
The sign line layout used throughout A bookkeeping device rather than a theorem; what carries the argument is the test value you evaluated in each piece.