Proofs of Derivative Applications Facts — Question 1

Question 1 - Solution

Since $f$ has a local maximum at $x=a$, there exists $\delta>0$ such that $$f(x)\le f(a) \quad\text{for all }x\text{ with }|x-a|<\delta.$$

Consider the difference quotient $$\frac{f(a+h)-f(a)}{h}$$ for small values of $h$.

Case 1: $h>0$

If $0<h<\delta$, then $a+h$ lies within the interval where $f(a)$ is a local maximum. Hence, $$f(a+h)-f(a)\le 0.$$

Since $h>0$, dividing by $h$ preserves the inequality: $$\frac{f(a+h)-f(a)}{h}\le 0.$$

Case 2: $h<0$

If $-\delta<h<0$, then again $a+h$ lies within the interval of the local maximum, so $$f(a+h)-f(a)\le 0.$$

Since $h<0$, dividing by $h$ reverses the inequality: $$\frac{f(a+h)-f(a)}{h}\ge 0.$$

Step 3: Take the limit

For $h>0$, the difference quotient is less than or equal to $0$. For $h<0$, the difference quotient is greater than or equal to $0$.

Since $f$ is differentiable at $a$, the limit $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$ exists and must be the same from both sides.

The only number that is both $\le 0$ and $\ge 0$ is $0$.

Therefore, $$f'(a)=0.$$

$$\boxed{f'(a)=0}$$