5.6 Applications of the mean value theorem

In the previous section, we interpreted the formula (5.18) from the mean value theorem in terms of a secant line and tangent line to the graph of $f$ . In many applications, however, it is useful to think about the mean value theorem in a different way.

MVT Interpretation 2: Comparing values.

If we rearrange (5.18), then we obtain

(5.20) (5.20)

f(b)-f(a)=f^{\prime}(c)(b-a).

Viewed in this way, the mean value theorem provides a useful tool for comparing two values $f(b)$ and $f(a)$ of a given function.

The first and second derivative tests

A simple way to compare the values $f(b)$ and $f(a)$ is to ask whether $f(b)>f(a)$ or $f(b)<f(a)$ . Moreover, we could ask whether $f$ is an increasing or a decreasing function. The rearranged mean value formula (5.20) relates this question to properties of the derivative.

Theorem 5.44 (First derivative test).

Suppose $a,b\in\mathbb{R}$ with $a<b$ . Let $f\colon[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$ .

1

If $f^{\prime}(x)>0$ for all $x\in(a,b)$ , then $f$ is strictly increasing on $[a,b]$ .
2

If $f^{\prime}(x)<0$ for all $x\in(a,b)$ , then $f$ is strictly decreasing on $[a,b]$ .
3

$f^{\prime}(x)=0$ for all $x\in(a,b)$ , then $f$ is constant on $[a,b]$ .

Exercise 5.45.

Apply the mean value theorem to prove Theorem 5.44.

Using Theorem 5.44, we can address a question left open in Section 5.5. Recall from Theorem 5.34 that any local maximum or local minimum of a differentiable function $f$ is a stationary point. We saw in Example 5.36 that the converse of Theorem 5.34 does not hold (that is, a stationary point need not be a local maximum or local minimum). However, we can often remedy this situation by considering information about second derivatives.

Theorem 5.46 (Second derivative test).

Let $I\subseteq\mathbb{R}$ be an open interval and $f\colon I\to\mathbb{R}$ be twice differentiable. Suppose $a\in I$ is a stationary point for $f$ (that is, $f^{\prime}(a)=0$ ).

1

If $f^{\prime\prime}(a)<0$ , then $a$ is a local maximum for $f$ .
2

If $f^{\prime\prime}(a)>0$ , then $a$ is a local minimum for $f$ .

Proof.

We shall only consider part 1), since the proof of part 2) is similar (or, alternatively, part 1) follows from part 2) by replacing $f$ with $-f$ ). By the definition of the second derivative and the hypothesis $f^{\prime}(a)=0$ , we have

f^{\prime\prime}(a)=\lim_{h\to 0}\frac{f^{\prime}(a+h)-f^{\prime}(a)}{h}=\lim_% {h\to 0}\frac{f^{\prime}(a+h)}{h}.

Since $f^{\prime\prime}(a)<0$ , by an argument similar to that used in Lemma 4.94 (see Exercise 5.47), there exists some $\delta>0$ such that $a+h\in I$ whenever $h\in(-\delta,\delta)$ and

(5.21) (5.21)

\frac{f^{\prime}(a+h)}{h}<0\qquad\text{for all $0<|h|<\delta$.}

If $h>0$ , then (5.21) implies that $f^{\prime}(a+h)<0$ . Thus, by the first derivative test from Theorem 5.44, the function $f$ is strictly decreasing on the interval $(a,a+\delta)$ . On the other hand, if $h<0$ , then (5.21) implies that $f^{\prime}(a+h)>0$ . Thus, by the first derivative test, $f$ is strictly increasing on the interval $(a-\delta,a)$ .

Since $f$ is differentiable, it is continuous by Theorem 5.20. By continuity and the monotonicity properties described above,

(5.22) (5.22)

f(a)=\lim_{u\to a}f(u)\geq f(x)\qquad\text{for all $x\in J:=(a-\delta,a+\delta% )$}

and so $a$ is a local maximum. ∎

Exercise 5.47.

Fill in the details of the proof of Theorem 5.46 as follows.

(i)

By unpacking the definition of $f^{\prime\prime}(a)$ and the definition of a limit of function, show there exists some $\delta>0$ such that (5.21) holds.
(ii)

Carefully justify the steps in (5.22).

Exercise 5.48.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be given by $f(x):=\sin x+\cos x$ for all $x\in\mathbb{R}$ . Show that $f$ has a local maximum at $\pi/4$ and a local minimum at $5\pi/4$ .

Note that Theorem 5.46 does not say anything about the case $f^{\prime\prime}(a)=0$ . Indeed, in this case the test is inconclusive.

Exercise 5.49.

For each part of the question, find a twice differentiable $f\colon\mathbb{R}\to\mathbb{R}$ satisfying $f^{\prime}(0)=f^{\prime\prime}(0)=0$ and the stated additional condition.

(i)

$f$ has neither a local maximum nor a local minimum at $0$ ;
(ii)

$f$ has a maximum at $0$ ;
(iii)

$f$ has a minimum at $0$ .

The mean value inequality

Another way to compare $f(b)$ and $f(a)$ is to obtain a bound on $|f(b)-f(a)|$ . In other words, we can ask how close the values $f(b)$ and $f(a)$ are to one another. The rearranged mean value formula (5.20) also relates this question to properties of the derivative.

Corollary 5.50 (Mean value inequality).

Suppose $a,b\in\mathbb{R}$ with $a<b$ . Let $f\colon[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$ . Suppose there exists some $M>0$ such that $|f^{\prime}(x)|\leq M$ for all $x\in(a,b)$ . Then

(5.23) (5.23)

|f(b)-f(a)|\leq M|b-a|.

Proof.

This is an immediate consequence of the mean value theorem. ∎

Remark 5.51.

If you are familiar with the physical interpretation of the derivative as the velocity of an object, then this provides useful intuition for Corollary 5.50. In particular, we think of $f(t)$ as describing the position of an object at time $t$ . We start at time $a$ and finish at time $b$ , so the quantity $|f(b)-f(a)|$ corresponds to the spatial displacement within that time window. The condition $|f^{\prime}(t)|\leq M$ tells us that the speed of our object is bounded above by $M$ . Thus,

|f(b)-f(a)|=\text{ displacement }\leq\text{distance}\leq\text{ maximum speed }% \times\text{ time taken }\leq M|b-a|,

which provides an intuitive explanation of (5.23). Note that this line of reasoning is not a mathematical proof, but it is helpful for understanding and interpreting the result.

Corollary 5.50 is an extremely useful result which pops up all over the place. Indeed, whenever you need to bound an expression of the form $f(b)-f(a)$ , the mean value inequality is a good tool to try first. Here we illustrate some typical applications.

Example 5.52.

The series $\sum_{k=1}^{\infty}\big{(}\sqrt{1+2^{-k}}-1\big{)}$ converges.

Proof.

We need to bound the size of the terms $\sqrt{1+2^{-k}}-1$ of the series. To do this, given $b>0$ we define a function $f\colon[0,b]\to\mathbb{R}$ by $f(x):=\sqrt{1+x}$ . Note that $f$ is continuous on $[0,b]$ . By Exercise 5.32 and the differentiation laws, we also know $f$ is differentiable on $(0,b)$ . Thus, the mean value theorem applies. In particular, there exists some $c\in(0,b)$ such that

f(b)-f(0)=f^{\prime}(c)(b-0).

Thus,

\sqrt{1+b}-1=\frac{b}{2\sqrt{1+c}}\leq b,

where the bound on the right-hand side follows because $c>0$ .

Substituting $b=2^{-k}$ into the above inequality, we see that

\sqrt{1+2^{-k}}-1\leq 2^{-k}\qquad\text{for all $k\in\mathbb{N}$.}

Since $\sum_{k=1}^{\infty}2^{-k}$ is a convergent geometric series, $\sum_{k=1}^{\infty}\big{(}\sqrt{1+2^{-k}}-1\big{)}$ converges by comparison. ∎

Exercise 5.53.

Use the mean value inequality to show that $\sum_{k=1}^{\infty}(1-\cos(1/k^{2}))$ converges.

Exercise 5.54.

Use the mean value inequality to give yet another proof of Bernoulli’s inequality

(1+x)^{n}\geq 1+nx\qquad\text{for all $x\in\mathbb{R}$ with $x\geq 0$ and all % $n\in\mathbb{N}$.}