Applications of the Mean Value Theorem

We say a function $f(x)$ is strictly increasing if whenever . $f(x)$ is strictly decreasing if implies . $f(x)$ is called monotone increasing if implies $f(a)\le f(b)$, and monotone decreasing if implies $f(a)\ge f(b)$.

Now suppose $f(x)$ is continuous for $a\le x\le b$ and f'(x) exists and is positive for all $x$ such that . Then let $c$ and $d$ be such that . Then surely $f(x)$ satisfies the hypotheses of the Mean Value Theorem on the interval $c\le x\le d$. Therefore

f(d) - f(c) = f'(\xi)(d - c), \quad \text{where } c < \xi < d .

But then f'(\xi) > 0, and . Therefore , or . This is exactly the condition that $f(x)$ be strictly increasing on $a\le x\le b$.

In similar fashion, one shows that
a) If f'(x) < 0 for , then $f(x)$ is strictly decreasing.
b) If f'(x) \ge 0 for , then $f(x)$ is monotone increasing.
c) If f'(x) \le 0 for , then $f(x)$ is monotone decreasing.

These are left as exercises. Observe, however, that there are functions which are, for instance, monotone decreasing, but which are not continuous, therefore do not even have derivatives throughout an interval. As an example, let

f(x) = \left\{ \begin{aligned} 1 - x, \quad 0 \le x < \frac{1}{2} \\ 0, \quad \frac{1}{2} \le x \le 1 \end{aligned} \right\} .

We have seen that the derivative of a constant is zero. Now we can prove the converse, namely that a continuous function $f(x)$ on $a\le x\le b$ whose derivative f'(x) = 0 for all , is a constant. For if , then f(d) - f(a) = f'(\xi)(d - a) for some , by the Mean Value Theorem. But f'(\xi) = 0, and $f(d)=f(a)$. Since $d$ could be any number of the interval, we have $f(x)=f(a)$ for all $x,a\le x\le b$, and $f(x)$ is a constant.

As a corollary to this result, let $f(x)$ and $g(x)$ be continuous on $a\le x\le b$ and let f'(x) = g'(x) for all . Then $f(x)-g(x)$ is continuous and has derivative 0, therefore is a constant $c$. So $f(x)=g(x)+c$. We have proved that two functions having equal derivatives in an interval differ by a constant function. (Note that it was the linearity of the operation of differentiation that made this an easy consequence of the preceding result.)

Next consider the function $f(x)=x-\sin x$. This is continuous and differentiable in any interval; its derivative is $1-\cos x\ge 0$. Therefore it is monotone increasing. But $f(0)=0$, so if $x\ge 0$, then $f(x)\ge f(0)=0$, and $x-\sin x\ge 0$ if $x\ge 0$. Likewise the function $-1+\frac{1}{2}{x}^2+\cos x$ has value 0 when $x=0$; its derivative is $x-\sin x\ge 0$ for $x\ge 0$ by the above. Hence it is also a monotone increasing function, and so $-1+\frac{1}{2}{x}^2+\cos x\ge 0$ when $x\ge 0$. Now

$$-x+\frac{x^3}{6}+\sin x\ge 0$$

for $x\ge 0$, by similar reasoning; continuing, we have

$$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cos x\ge 0,$$$$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\sin x\ge 0,$$$$-1+\frac{x^2}{2!}-\frac{x^4}{4!}+\frac{x^6}{6!}+\cos x\ge 0,$$$$⋮$$

for all $x\ge 0$, where $n!$ ($𝒏$-factorial) is the product $1\cdot 2\cdot \cdots \cdot n$. (Note that $2!=2,1!=1$. We agree to write $0!=1$.) These can be transposed to give the following inequalities for all $x\ge 0$:

$$\sin x\le x$$$$1-\frac{1}{2!}{x}^2\le \cos x\le 1$$$$x-\frac{x^3}{3!}\le \sin x\le x$$$$1-\frac{x^2}{2!}\le \cos x\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}$$$$x-\frac{x^3}{3!}\le \sin x\le x-\frac{x^3}{3!}+\frac{x^5}{5!}$$

Thus we have polynomials for approximating $\sin x$ and $\cos x$, at least for $x\ge 0$ (one can also apply the method to the case $x\le 0$). The error in using one of these polynomials as an approximation is at most the difference between it and the next polynomial in the sequence; for example, if we use $1-\frac{1}{3!}=.833\dots$ to estimate $\sin 1$, our error is at most $\frac{1}{5!}=\frac{1}{120}$.

PROBLEMS