MA005: Calculus I

Geometrically, the Mean Value Theorem is a "tilted" version of Rolle's Theorem (Fig. 5). In each theorem we conclude that there is a number $\mathrm{c}$ so that the slope of the tangent line to $\mathrm{f}$ at $\mathrm{x}=\mathrm{c}$ is the same as the slope of the line connecting the two ends of the graph of $\mathrm{f}$ on the interval $[\mathrm{a}, \mathrm{b}]$. In Rolle's Theorem, the two ends of the graph of $\mathrm{f}$ are at the same height, $\mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{b})$, so the slope of the line connecting the ends is zero. In the Mean Value Theorem, the two ends of the graph of $\mathrm{f}$ do not have to be at the same height so the line through the two ends does not have to have a slope of zero.

Fig. 5

Proof: The proof of the Mean Value Theorem uses a tactic common in mathematics: introduce a new function which satisfies the hypotheses of some theorem we already know and then use the conclusion of that previously proven theorem. For the Mean Value Theorem we introduce a new function, $\mathrm{h}(\mathrm{x})$, which satisfies the hypotheses of Rolle's Theorem. Then we can be certain that the conclusion of Rolle's Theorem is true for $\mathrm{h}(\mathrm{x})$, and the Mean Value Theorem for $\mathrm{f}$ follows from the conclusion of Rolle's Theorem for $\mathrm{h}$.

First, let $\mathrm{g}(\mathrm{x})$ be the straight line through the ends $(\mathrm{a}, \mathrm{f}(\mathrm{a}))$ and $(\mathrm{b}, \mathrm{f}(\mathrm{b}))$ of the graph of $\mathrm{f}$. The function $\mathrm{g}$ goes through the point $(\mathrm{a}, \mathrm{f}(\mathrm{a}))$ so $\mathrm{g}(\mathrm{a})=\mathrm{f}(\mathrm{a})$. Similarly, $\mathrm{g}(\mathrm{b})=\mathrm{f}(\mathrm{b})$. The slope of the linear function $g$ is $\frac{f(b)-f(a)}{b-a}$ so $g^{\prime}(x)=\frac{f(b)-f(a)}{b-a}$ for all $x$ between $a$ and $b$, and $g$ is continuous and differentiable. (The formula for $\mathrm{g}$ is $\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{a})+\mathrm{m}(\mathrm{x}-\mathrm{a})$ with $\mathrm{m}=(\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})) /(\mathrm{b}-\mathrm{a})$.)

Define $h(x)=f(x)-g(x)$ for $a \leq x \leq b$ (Fig. 6). The function $h$ satisfies the hypotheses of Rolle's theorem:
$\mathrm{h}(\mathrm{a})=\mathrm{f}(\mathrm{a})-\mathrm{g}(\mathrm{a})=0$ and $\mathrm{h}(\mathrm{b})=\mathrm{f}(\mathrm{b})-\mathrm{g}(\mathrm{b})=0$
$\mathrm{h}(\mathrm{x})$ is continuous for $\mathrm{a} \leq \mathrm{x} \leq \mathrm{b}$ since both $\mathrm{f}$ and $\mathrm{g}$ are continuous there, and $\mathrm{h}(\mathrm{x})$ is differentiable for $\mathrm{a} < \mathrm{x} < \mathrm{b}$ since both $\mathrm{f}$ and $\mathrm{g}$ are differentiable there, so the conclusion of Rolle's Theorem applies to $\mathrm{h}$: there is a $c$, between $a$ and $b$, so that $h^{\prime}(c)=0$.

Fig. 6

The derivative of $h(x)=f(x)-g(x)$ is $h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)$ so we know that there is a number $c$, between $a$ and $b$, with $h^{\prime}(c)=0$. But $0=h^{\prime}(c)=f^{\prime}(c)-g^{\prime}(c)$ so $f^{\prime}(c)=g^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$.

Graphically, the Mean Value Theorem says that there is at least one point $\mathrm{c}$ where the slope of the tangent line, $f^{\prime}(c)$, equals the slope of the line through the end points of the graph segment, $(\mathrm{a}, \mathrm{f}(\mathrm{a}))$ and $(\mathrm{b}, \mathrm{f}(\mathrm{b}))$. Fig. 7 shows the locations of the parallel tangent lines for several functions and intervals.

Fig. 7

The Mean Value Theorem also has a very natural interpretation if $\mathrm{f}(\mathrm{x})$ represents the position of an object at time $x : f^{\prime}(x)$ represents the velocity of the object at the instant $x$, and $\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}=\frac{\text { change in position }}{\text { change in time }}$ represents the average (mean) velocity of the object during the time interval from time $a$ to time $b$. The Mean Value Theorem says that there is a time $c$, between $a$ and $b$, when the instantaneous velocity, $f^{\prime}(c)$, is equal to the average velocity for the entire trip, $\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}$. If your average velocity during a trip is $30$ miles per hour, then at some instant during the trip you were traveling exactly $30$ miles per hour.

Practice 2: For $f(x)=5 x^{2}-4 x+3$ on the interval $[1,3]$, calculate $m=\frac{f(b)-f(a)}{b-a}$ and find the value of $c$ so that $f^{\prime}(c)=m$.