I need to prove these two results:
- $f(a) = 0$ and $f'(a)\neq 0$. You can use the definition of the derivative to show the left and right derivatives of $|f|$ at $a$ are unequal.
- $f(a) = 0$ and $f'(a) = 0$. You can use the definition of the derivative to show that $|f|$ is differentiable at $a$ with $|f|'(a) =0$.
EDIT- My attempt: For the second one, I tried to use what was given. By the definition of derivative of $f$ at the point $a$, we have: $$ f'(a) = \lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}$$ It gives me from the given data: $$ 0 = \lim\limits_{x \to a} \frac{f(x)}{x-a}$$ Let $g(x)= \frac{f(x)}{x-a}$. It is easy to observe that: $g(x)\to 0 \Leftrightarrow |g(x)|\to 0$. So, $$ \lim\limits_{x \to a} |g(x)|= 0\lim\limits_{x \to a} \frac{|f(x)|}{|x-a|} = 0 $$ By definition of absolute value: $$ |g(x)| = \begin{cases} \frac{|f(x)|}{x-a}, & \text{if $x \gt a$} \\[2ex] \frac{|f(x)|}{-(x-a)}, & \text{if $x \lt a$} \end{cases}$$ So the right-hand derivative of $|f(x)|$ at $x=a$ is: $$ \lim\limits_{x \to a^+} \frac{|f(x)|}{x-a} = 0 $$ and the left-hand derivative of $|f(x)|$ at $x=a$ is: $$ \lim\limits_{x \to a^-} \frac{|f(x)|}{x-a} = 0 $$ $$ \therefore |f'(a)| = \lim\limits_{x \to a} \frac{|f(x)|-|f(a)|}{(x-a)} = 0 $$
Please suggest if my solution works.
Assume that $$f(a)=0\;\;and\;\;f'(a)>0$$
then
$$\lim_{x\to a^+}\frac{f(x)}{x-a}>0 \implies$$
$$\exists \eta>0 :\forall x\in(a,a+\eta) \frac{f(x)}{x-a}>0$$ $$\implies \forall x\in(a,a+\eta)\;\; f(x)>0$$
$$\implies \lim_{x\to a^+}\frac{|f(x)|}{x-a}=f'(a)$$
By the same, we will have
$$\lim_{x\to a^-}\frac{|f(x)|}{x-a}=-f'(a)$$