Q. 69 - Calculus

Question

Use Rolle’s Theorem to prove the slightly more general theorem from Exercise 68: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f\left(a\right)=f\left(b\right)$, then there is some value $c\in (a,b)$ with $f\left(c\right)=0$.

Step-by-Step Solution

Answer

Rolle's Theorem proves that the following theorem holds true.

Step 1. Given Information.

$f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

Step 2. Rolle's Theorem.

Here, $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Then,

$$\begin{gathered} g\left(a\right)=f\left(a\right)-f\left(a\right) \\ =0 \end{gathered}$$

$$\begin{gathered} g\left(b\right)=f\left(b\right)-f\left(b\right) \\ =0 \end{gathered}$$

Hence, Rolle's Theorem applies to the function $g\left(x\right)$. Therefore, we can conclude that there is some $c\in (a,b)$ such that $g\text{'}\left(c\right)=0$.

$$\begin{gathered} g\text{'}\left(x\right)=f\text{'}\left(x\right)-0 \\ =f\text{'}\left(x\right) \end{gathered}$$

This also means that $f\text{'}\left(c\right)=0$ as desired.