Q. 68 - Calculus

Question

Follow the method of proof that we used for Rolle’s Theorem to prove the following slightly more general theorem: 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

The given theorem is proved by Rolle's Theorem.

Step 1. Given Information.

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

Step 2. Using Rolle's Theorem.

We have to show that if$f\left(a\right)=f\left(b\right)$, then there is some value $c\in (a,b)$ with $f\text{'}\left(c\right)=0$.

Let us consider the Extreme Value Theorem if a function $f\left(x\right)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Then there exists at least one value $c\in (a,b)$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ as,

$$\begin{gathered} f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a} \\ =\frac{f\left(b\right)-f\left(b\right)}{b-a} \\ =\frac{0}{b-a} \\ =0 \end{gathered}$$

Hence, if $f\left(a\right)=f\left(b\right)$ then exists $c\in (a,b)$ with $f\text{'}\left(c\right)=0$.