05Linear approximation, differentials, extreme values, and the Mean Value Theorem

You are being asked whether the converse of a theorem is also true. Try to remember a graph with a flat spot that is not a peak or a valley.

Fermat's theorem says local extremum plus differentiability implies $f'(c)=0$. Nothing in it licenses the reverse reading.

Take $f(x)=x^{3}$ and $c=0$: $f'(x)=3x^{2}$, so $f'(0)=0$, while $f(x)<0$ for $x<0$ and $f(x)>0$ for $x>0$.

So the statement is false; $x^{3}$ at $0$ is the standard counterexample.

An absolute minimum on a closed interval is either at an endpoint or strictly inside. Ask what an interior absolute minimum automatically is.

An interior absolute minimum is in particular a local minimum, so it has to be one of the local extrema you were told about.

The candidates are therefore $f(1)=3$, the local minimum $1$ at $x=5$, and $f(6)=3$; the local maximum $8$ at $x=2$ cannot be the smallest.

The smallest of $3$, $1$ and $3$ is $1$, so the absolute minimum value is $1$ at $x=5$.

Check the two hypotheses on each line separately: is the function continuous at every point of the set, and is the set a closed bounded interval.

The guarantee needs continuity on a closed and bounded interval, and it needs both at once.

On $[0,\pi/4]$ the tangent function is continuous, since the trouble at $\pi/2$ is far outside, and the interval is closed with finite ends.

So the first line is the one covered by the theorem.

The line needs two numbers from the base point, and then the step $x-a$ does the rest.

$L(x)=f(a)+f'(a)(x-a)$, and for $f(x)=x^{-1}$ the derivative is $f'(x)=-x^{-2}$.

With $a=4$: $f(4)=0.25$ and $f'(4)=-\dfrac{1}{16}=-0.0625$; the target $4.1$ gives a step of $x-a=0.1$.

So $L(4.1)=0.25-0.00625=0.24375$, and the true value is $0.243902\ldots$

The tolerance is a change in the input, so you want the change in the output that the tangent predicts.

$dV=\dfrac{dV}{dr}\,dr$, and differentiating $\dfrac43\pi r^{3}$ with respect to $r$ gives $4\pi r^{2}$.

With $r=6$ and $dr=0.1$: $dV=4\pi(6)^{2}(0.1)=14.4\pi$, and $V=\tfrac43\pi(216)=288\pi$.

So $dV=14.4\pi\approx 45.2$ cm³, which is $14.4/288=5\%$ of the volume.

Same four steps as always. Expect the symmetry of the function to show up in the answer.

Candidates are the critical numbers inside $(-2,2)$ together with the endpoints; differentiate and factor.

With $f(x)=x^{4}-2x^{2}+3$: $f'(x)=4x^{3}-4x=4x(x-1)(x+1)$, so the critical numbers are $-1$, $0$, $1$, all inside, and the endpoints are $\pm2$.

The values are $f(\pm2)=11$, $f(\pm1)=2$, $f(0)=3$, so the maximum is $11$ and the minimum is $2$.

Two searches as always: where $f'$ is zero, and where $f'$ fails to exist. Look at the denominator before you start.

Quotient rule: $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}$, and a fraction is zero exactly when its numerator is.

With $u=x$ and $v=x^{2}+1$: $f'(x)=\dfrac{(1)(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}$.

The numerator vanishes at $x=\pm1$, and the denominator never does, so those two are the complete list.

Check the two hypotheses one at a time and notice which interval each one is about.

Continuity is required on $[0,4]$ and differentiability only on $(0,4)$; the conclusion is $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.

Here $f(4)=2$ and $f(0)=0$, so the average rate is $\dfrac{2-0}{4-0}=\dfrac12$, and $f'(x)=\dfrac{1}{2\sqrt{x}}$.

Solving $\dfrac{1}{2\sqrt c}=\dfrac12$ gives $\sqrt c=1$, so $c=1$.

The usual four steps. Expect one critical number inside, and remember that both endpoints are candidates.

Differentiate and solve $f'(x)=0$; then compare values of $f$, not of $f'$.

Here $f'(x)=\cos x-\sin x=0$ gives $\tan x=1$, so $x=\pi/4$; the candidates are $0$, $\pi/4$ and $\pi/2$.

The values are $f(0)=1$, $f(\pi/4)=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2$ and $f(\pi/2)=1$, so the maximum is $\sqrt2$.

Read the four steps against the definition of the linearization, one symbol at a time. Also ask whether the final number is even plausible.

In $L(x)=f(a)+f'(a)(x-a)$ the last factor is the step away from the base point, not the input itself.

With $a=8$ and $x=8.06$ the step is $x-a=0.06$, so the correct evaluation is $L(8.06)=2+\tfrac{1}{12}(0.06)$.

That gives $2.005$, so the error is in Step 4.

You know something about $f'$ everywhere and something about $f$ at one point, and you want $f$ at another point. Which theorem connects those three things?

The Mean Value Theorem gives $f(6)-f(2)=f'(c)(6-2)$ for some $c$ between $2$ and $6$; you do not need to know which $c$.

With $f(2)=5$ and $f'\le 3$ everywhere: $f(6)-5=4f'(c)\le 4(3)=12$.

So $f(6)\le 17$, and the line $f(x)=5+3(x-2)$ reaches exactly $17$.

Two questions, two different unknowns: one is a change, the other is a rate. Read what each part gives you before choosing a tool.

For a change caused by a small step, $dV=\dfrac{dV}{dr}\,dr$; for a rate in time, differentiate the same relation with respect to $t$ to get $\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$.

Both use $\dfrac{dV}{dr}=4\pi r^{2}$, which is $4\pi(25)=100\pi$ at $r=5$. Part (a) has $dr=0.02$ cm; part (b) has $\dfrac{dr}{dt}=0.3$ cm/s.

So $dV=100\pi(0.02)=2\pi\approx 6.3$ cm³ and $\dfrac{dV}{dt}=100\pi(0.3)=30\pi\approx 94.2$ cm³/s.

Differentiating this needs two rules at once. Decide which is the outer structure before writing anything.

It is a product, and the second factor needs the chain rule: $\left(\sqrt{u}\right)'=\dfrac{u'}{2\sqrt u}$.

With $f(x)=x\sqrt{4-x^{2}}$: $f'(x)=\sqrt{4-x^{2}}+x\cdot\dfrac{-2x}{2\sqrt{4-x^{2}}}=\dfrac{4-2x^{2}}{\sqrt{4-x^{2}}}$.

The numerator vanishes at $x=\pm\sqrt2$, and $f(\pm\sqrt2)=\pm 2$ while $f(\pm 2)=0$.

Check the hypotheses of the guarantee first, at every point of the interval, including the awkward one.

The Extreme Value Theorem needs continuity at every point of a closed bounded interval; at $x=2$ compare the limit with the value.

Here $f(2)=0$ while $f(x)\to+\infty$ as $x\to 2^{+}$ and $f(x)\to-\infty$ as $x\to 2^{-}$, so $f$ is defined but not continuous at $2$.

Values climb past every ceiling just right of $2$ and fall below every floor just left of it, so neither extreme exists.