The Mean Value Theorem and Its Consequences

If the Mean Value Theorem was just an isolated result about the existence of a particular point $\mathrm{c}$, it would not be very important or useful. However, the Mean Value Theorem is the basis of several results about the behavior of functions over entire intervals, and it is these consequences which give it an important place in calculus for both theoretical and applied uses.

The next two corollaries are just the first of many results which follow from the Mean Value Theorem.

We already know, from the Main Differentiation Theorem, that the derivative of a constant function. $f(x)=k$ is always $0$, but can a nonconstant function have a derivative which is always $0$? The first corollary says no.

Proof: Assume $\mathrm{f}^{\prime}(\mathrm{x})=0$ for all $\mathrm{x}$ in an interval $\mathrm{I}$, and pick any two points $a$ and $\mathrm{b}(\mathrm{a} \neq \mathrm{b})$ in the interval. Then, by the Mean Value Theorem, there is a number $c$ between $a$ and $b$ so that $\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}$. By our assumption, $\mathrm{f}^{\prime}(\mathrm{x})=0$ for all $\mathrm{x}$ in $\mathrm{I}$ so we know that $0=\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}$ and we can conclude that $\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})=0$ and $\mathrm{f}(\mathrm{b})=\mathrm{f}(\mathrm{a})$. But $a$ and $\mathrm{b}$ were any two points in $I$, so the value of $f(x)$ is the same for any two values of $x$ in $I$, and $f$ is a constant function on the interval $I$.

We already know that if two functions are parallel (differ by a constant), then their derivatives are equal, but can two nonparallel functions have the same derivative? The second corollary says no.

Proof: This corollary involves two functions instead of just one, but we can imitate the proof of the Mean Value Theorem and introduce a new function $\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})$. The function $\mathrm{h}$ is differentiable, and $h^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)=0$ for all $x$ in $I$, so, by Corollary $1, h(x)$ is a constant function and $\mathrm{K}=\mathrm{h}(\mathrm{x})=\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})$ for all $\mathrm{x}$ in the interval. Then $\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{K}$.

We will use Corollary 2 hundreds of times in Chapters 4 and 5 when we work with "integrals". Typically you will be given the derivative of a function, $f^{\prime}(x)$, and asked to find all functions $f$ which have that derivative. Corollary 2 tells us that if we can find one function $\mathrm{f}$ which has the derivative we want, then the only other functions which have the same derivative are $\mathrm{f}(\mathrm{x})+\mathrm{K}$: once you find one function with the right derivative, you have essentially found all of them.

Example 2: (a) Find all functions whose derivatives equal $2 \mathrm{x}$.
(b) Find a function $g(x)$ with $g^{\prime}(x)=2 x$ and $g(3)=5$.

Solution: (a) We can recognize that if $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$ then $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}$ so one function with the derivative we want is $f(x)=x^{2}$. Corollary 2 guarantees that every function $g$ whose derivative is $2 x$ has the form $\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{K}=\mathrm{x}^{2}+\mathrm{K}$. The only functions with derivative $2 \mathrm{x}$ have the form $\mathrm{x}^{2}+\mathrm{K}$.

(b) Since $g^{\prime}(x)=2 x$, we know that $g$ must have the form $g(x)=x^{2}+K$, but this is a whole "family" of functions (Fig. 8), and we want to find one member of the family . We know that $g(3)=5$ so we want to find the member of the family which goes through the point $(3,5)$. All we need to do is replace the $\mathrm{g}(\mathrm{x})$ with 5 and the $\mathrm{x}$ with 3 in the formula $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{K}$, and then solve for the value of $\mathrm{K}: 5=\mathrm{g}(3)=(3)^{2}+\mathrm{K}$ so $K=-4$. The function we want is $g(x)=x^{2}-4$.

Fig. 8

Practice 3: Restate Corollary 2 as a statement about the positions and velocities of two cars.