differentiation of $g(x) = \lvert f(x)\rvert$ where $f(x)$ and $D(f(x)) = 0$

Question

I'm really stumped on this problem and don't know how to go about it. It says $g(x)$ = $|f(x)|$ and to show that if $f(c) = 0$ and g is differentiable at c, then one must have $D(f)(c) = 0$. Everywhere I look on the internet it says that |x| is not differentiable at 0 even there is nothing on absolute value of functions.

I tried doing something like $$\lim_{x\to\ c^+} \frac{g(x) - g(c)}{x-c} = \lim_{x\to\ c^-} \frac{g(x) - g(c)}{x-c}$$

but didn't know where to go afterwards.

Also tried doing $|x| = \sqrt{x^2}$ but that did not work either.

I'm thinking of whether I should be looking more into maximas or something.

What do you guys think? Thanks

Answer 1

$g(x) \geq 0$ for all $x$ and $g(c)=0$, so if $g$ is differentiable at $c$ there is a local minimum at $c$ and therefore $g'(c)=0$

now $$ \dfrac{g(c+h)-g(c)}{h}=\dfrac{g(c+h)}{h}=\dfrac{|f(c+h)|}{h} \rightarrow 0 \text{ for } h\rightarrow 0 $$ and note that (for example) a sequence $a_n \rightarrow 0$ iff $|a_n|\rightarrow 0$

Answer 2

Generally it is not true. If $\chi_{\mathbb{Q}} $ is a charakteristic function of rationals then the function $g(t) =\left|\chi_{\mathbb{Q}} (t) -\frac{1}{2} \right|t^2 $ is differentiable everywhwre but the function $f(t) =\left(\chi_{\mathbb{Q}} (t) -\frac{1}{2} \right)t^2$ is equal zero only at zero.