The Mean Value Theorem and Its Consequences

Suppose we pick any two points on the $x$-axis and think about all of the differentiable functions which go through those two points (Fig. 1).

Fig. 1

Since our functions are differentiable, they must be continuous and their graphs can not have any holes or breaks. Also, since these functions are differentiable, their derivatives are defined everywhere between our two points and their graphs can not have any "corners" or vertical tangents. The graphs of the functions in Fig. 1 can still have all sorts of shapes, and it may seem unlikely that they have any common properties other than the ones we have stated, but Michel Rolle found one. He noticed that every one of these functions has one or more points where the tangent line is horizontal (Fig. 2), and this result is named after him.

Fig. 2

Proof: We consider two cases: when $f(x)=f(a)$ for all $x$ in $(a, b)$ and when $f(x) \neq f(a)$ for some $x$ in $(a, b)$.

Case I, $\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})$ for all $\mathrm{x}$ in $(\mathrm{a}, \mathrm{b})$: If $\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})$ for all $\mathrm{x}$ between $\mathrm{a}$ and $\mathrm{b}$, then $\mathrm{f}$ is a horizontal line segment and $\mathrm{f}^{\prime}(\mathrm{c})=0$ for all values of $\mathrm{c}$ strictly between $\mathrm{a}$ and $\mathrm{b}$.

Case II, $f(x) \neq f(a)$ for some $x$ in $(a, b)$: Since $f$ is continuous on the closed interval $[a, b]$, we know from the Extreme Value Theorem that $\mathrm{f}$ must have a maximum value in the closed interval $[\mathrm{a}, \mathrm{b}]$ and a minimum value in the interval.

If $f(x)>f(a)$ for some value of $x$ in $[a, b]$, then the maximum of $f$ must occur at some value $c$ strictly between $a$ and $b, a < c < b$. (Why can't the maximum be at $a$ or $b$?) Since $f(c)$ is a local maximum of $f$, then $c$ is a critical number of $f$ and $f^{\prime}(c)=0$ or $f^{\prime}(c)$ is undefined. But $f$ is differentiable at all $\mathrm{x}$ between $\mathrm{a}$ and $\mathrm{b}$, so the only possibility left is that $\mathrm{f}^{\prime}(\mathrm{c})=0$.

If $f(x) < f(a)$ for some value of $x$ in $[a, b]$, then $f$ has a minimum at some value $x=c$ strictly between a and $b$, and $f^{\prime}(c)=0$.

In either case, there is at least one value of $\mathrm{c}$ between $a$ and $\mathrm{b}$ so that $\mathrm{f}^{\prime}(\mathrm{c})=0$.

Example 1: Show that $f(x)=x^{3}-6 x^{2}+9 x+2$ satisfies the hypotheses of Rolle's Theorem on the interval $[0,3]$ and find the value of $c$ which the theorem says exists.

Solution: $\mathrm{f}$ is a polynomial so it is continuous and differentiable everywhere. $\mathrm{f}(0)=2$ and $f(3)=2$. $f^{\prime}(x)=3 x^{2}-12 x+9=3(x-1)(x-3)$ so $f^{\prime}(x)=0$ at $1$ and $3$.
The value $c=1$ is between $0$ and $3$. Fig. 3 shows the graph of $f$.

Fig. 3

Practice 1: Find the value(s) of c for Rolle's Theorem for the functions in Fig. 4.

Fig. 4