12.4 The Mean Value Theorem

12.24   Lemma (Rolle's Theorem) Let $a,b$ be real numbers with $a<b$ and let $f\colon [a,b]\to\mbox{{\bf R}}$ be a function that is continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose that $f(a)=f(b)$. Then there is a point $c\in (a,b)$ such that $f^\prime (c)=0$.

Proof: By the extreme value property, $f$ has a maximum at some point $A\in [a,b]$. If $A\in (a,b)$, then $f^\prime (A)=0$ by the critical point theorem. Suppose $A\in \{a,b\}$. By the extreme value property, $f$ has a minimum at some point $B\in [a,b]$. If $B\in (a,b)$ then $f^\prime (B)=0$ by the critical point theorem. If $B\in \{a,b\}$, then we have $\{A,B\}\subset \{a,b\}$ so $f(A)=f(B)=f(a)=f(b)$. Hence in this case the maximum value and the minimum value taken by $f$ are equal, so $f(x)=f(a)$ for $x\in [a,b]$ so $f^\prime (x)=0$ for all $x\in (a,b)$. $\diamondsuit$

Rolle's theorem is named after Michel Rolle (1652-1719). An English translation of Rolle's original statement and proof of the theorem can be found in [43, pages 253-260]. It takes a considerable effort to see any relation between what Rolle says and what our form of Rolle's theorem says.

12.25   Theorem (Mean value theorem.) Let $a,b$ be real numbers and let $f\colon [a,b]\to\mbox{{\bf R}}$ be a function that is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there is a point $c\in (a,b)$ such that $${f^\prime (c)={{f(b)-f(a)}\over {b-a}}}$$; i.e., there is a point $c$ where the slope of the tangent line is equal to the slope of the line joining $\Big( a,f(a)\Big)$ to $\Big(b,f(b)\Big)$.

Proof: The equation of the line joining $\Big( a,f(a)\Big)$ to $\Big(b,f(a)\Big)$ is

$$y=l(x)=f(a)+{{f(b)-f(a)}\over {b-a}} (x-a).$$

Let

$$\begin{eqnarray*} F(x)&=& f(x)-l(x)\\ &=& f(x)-f(a)-{{f(b)-f(a)}\over {b-a}}(x-a).\\ \end{eqnarray*}$$

Then $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $F(a)=F(b)=0$. By Rolle's theorem there is a point $c\in (a,b)$ where $F^\prime (c)=0$.

Now

$$F^\prime (x)=f^\prime (x)-{{f(b)-f(a)}\over {b-a}},$$

so

$$\begin{eqnarray*} F^\prime (c)=0 &\mbox{$\Longrightarrow$}& f^\prime (c)-{{f(b)-... ...f^\prime (c)={{f(b)-f(a)}\over {b-a}}.\mbox{ $\diamondsuit$}\cr \end{eqnarray*}$$

12.26   Corollary. Let $J$ be an interval in $\mbox{{\bf R}}$ and let $f\colon J\to\mbox{{\bf R}}$ be a function that is continuous on $J$ and differentiable at the interior points of $J$. Then

$$\begin{eqnarray*} f^\prime (x) =0 \mbox{ for all } x\in {\rm interior\;} (J) & \... ...mbox{$\Longrightarrow$}& f \mbox{ is strictly increasing on } J. \end{eqnarray*}$$

Proof: I will prove the second assertion. Suppose $f^\prime (x)\leq 0$ for all $x\in\mbox{interior} (J)$. Let $s,t$ be points in $J$ with $s<t$. Then by the mean value theorem

$$f(t)-f(s)=f^\prime (c)(t-s) \mbox{ for some } c\in (s,t).$$

Since $f^\prime (c)\leq 0$ and $(t-s)>0$, we have $f(t)-f(s)=f^\prime (c)(t-s)\leq 0$; i.e., $f(t)\leq f(s)$. Thus $f$ is decreasing on $J.\mbox{ $\diamondsuit$}$

12.27   Exercise. A Prove the first assertion of the previous corollary; i.e., prove that if $f$ is continuous on an interval $J$, and $f^\prime (x)=0$ for all $x\in\mbox{interior} (J)$, then $f$ is constant on $J$.

12.28   Definition (Antiderivative) Let $f$ be a real valued function with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$. Let $J$ be an interval such that $J\subset\mbox{{\rm dom}}(f)$. A function $F$ is an antiderivative for $f$ on $J$ if $F$ is continuous on $J$ and $F^\prime (x)=f(x)$ for all $x$ in the interior of $J$.

12.29   Examples. Since $${{ d \over dx} (x^3 + 4) = 3x^2}$$, we see that $x^3 + 4$ is an antiderivative for $3x^2$. Since

$${d\over dx} (\cos^2(x)) = 2 \cos(x) (-\sin(x)) = -2\sin(x)\cos(x),$$

and

$${d\over dx} (-\sin^2(x)) = - 2 \cdot \sin(x) \cos(x)$$

we see that $\cos^2$ and $-\sin^2$ are both antiderivatives for $-2\sin\cdot\cos$.

We will consider the problem of finding antiderivatives in chapter 17. Now I just want to make the following observation:

12.30   Theorem (Antiderivative theorem.) Let $f$ be a real valued function with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$ and let $J$ be an interval with $J\subset\mbox{{\rm dom}}(f)$. If $F$ and $G$ are two antiderivatives for $f$ on $J$, then there is a number $c \in \mbox{{\bf R}}$ such that

$$F(x)=G(x)+c \mbox{ for all } x\in J.$$

12.31   Exercise. A Prove the antiderivative theorem.

12.32   Definition (Even and odd functions.) A subset $S$ of $\mbox{{\bf R}}$ is called symmetric if $(x\in S\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}-x\in S)$. A function $f$ is said to be even if $\mbox{{\rm dom}}(f)$ is a symmetric subset of $\mbox{{\bf R}}$ and

$$f(x)=f(-x) \mbox{ for all } x\in\mbox{{\rm dom}}(f),$$

and $f$ is said to be odd if $\mbox{{\rm dom}}(f)$ is a symmetric subset of $\mbox{{\bf R}}$ and

$$f(x)=-f(-x)\mbox{ for all } x\in\mbox{{\rm dom}}(f)$$

.

12.33   Example. If $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ and $f(x)=x^n$, then $f$ is even if $n$ is even, and $f$ is odd if $n$ is odd. Also $\cos$ is an even function and $\sin$ is an odd function, while $\ln$ is neither even or odd.

12.34   Example. If $f$ is even, then $V\Big({\rm graph} (f)\Big)={\rm graph} (f)$ where $V$ is the reflection about the vertical axis. If $f$ is odd, then $R_\pi \Big({\rm graph}(f)\Big)={\rm graph}(f)$ where $R_\pi$ is a rotation by $\pi$ about the origin.

12.35   Exercise. A Are there any functions that are both even and odd?

12.36   Exercise. A

a)

If $f$ is an arbitrary even differentiable function, show that the derivative of $f$ is odd.

b)

If $g$ is an arbitrary odd differentiable function, show that the derivative of $g$ is even.