Question

Prove part (c) of Theorem $3.6$: Suppose f is differentiable on an interval I; if f is zero on the interior of I, then f is constant on I. (Hint: use the Mean Value Theorem to show that any two numbers a and b in I must be equal.)

Short answer

The part(c) of theorem$3.6$ is proved.

Step-by-step solution

Step 1. Given Information 

We are given that f is differentiable on an interval I.

Step 2. Proving the statement 

Let f be a differentiable function on an interval I and its derivative f' is zero inside I.

The objective is to prove that f is constant on I.

Suppose that $a,b\in I$and $b>a$.

By the definition of a constant function, $f\left(b\right)=f\left(a\right)$

As f is differentiable, it is continuous on the interval.

As $[a,b]$ is contained in the interval I, f satisfies the hypotheses of the Mean Value Theorem on$[a,b]$,

Therefore, it can be concluded that there exists some $c\in (a,b)$such that,

$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\dots ..\left(1\right)$

Step 3. Proving the statement 

Rewriting the equation as, we get

$f\left(b\right)-f\left(a\right)=f^{\text{'}}\left(c\right)(b-a)$

Since $c\in (a,b)$, it follows that c is in the interior of I, and thus by hypothesis $f\text{'}\left(c\right)=0$.

The condition $b>a$implies that $b-a>0$

Therefore, $f\left(b\right)-f\left(a\right)$is the product of a positive number and a zero, which should be zero.

Thercfore, a differentiable function f on an interval l, if f' is zero on the interior of l, then f is constant on I.