Prove part (b) of Theorem $3.6$: Suppose f is differentiable on an interval I; if f' is negative on the interior ofI, then f is decreasing on I.

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

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

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

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

By the definition of decreasing 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,

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

To show that $f\left(b\right)<f\left(a\right)$, it suffices to show that $f\left(b\right)-f\left(a\right)<0$,

Rewrite the equation $\left(1\right)$as,

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

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 $b-a>0$

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

Hence, a differentiable function f on an interval l, if f' is negative on the interior of I, then f is decreasing on I .