Absolutely true

If $f(x) = 0$ then $f'(x) = 0$ .

Suppose that $f'(x) = k > 0$ . By definition, for any $0 < \epsilon < k$ there exists $c < x < d$ such that for any $\xi \in \langle c,d\rangle$ , $k - \epsilon < f(\xi)/(\xi-x) < k + \epsilon$ .

But then for $c < \xi < x$ we have $f(\xi)$ is negative and for $x < \xi < d$ , $f(\xi)$ is positive. Now consider the function $g(x) = |f(x)|$ on that same interval. Then for $c < \xi < x$ , we find that $g(\xi)/(\xi-x) = -f(\xi)/(\xi-x)$ lies between $-k - \epsilon$ and $-k + \epsilon$ , while for $x < \xi < d$ , $g(\xi)/(\xi -x) = f(\xi)/(\xi-x)$ lies between $k - \epsilon$ and $k + \epsilon$ . Thus $\lim_{\xi \uparrow x} \frac{g(\xi)}{\xi-x} = -k\ \ \ \ \ \ \ \ \ \lim_{\xi \downarrow x} \frac{g(\xi)}{\xi-x} = +k,$ showing that the derivative $g'(x)$ is not defined. A similar conclusion follows for $k < 0$ . Therefore, in order for the derivative $g'(x)$ to be defined, we must have $f'(x) = k = 0$ .

To prove the other three statements wrong, consider the function $f(x) = x^3$ . It has a zero at $x = 0$ but nowhere else; the absolute value function $g(x) = |x^3|$ has derivative $g'(x) = \pm 3x^2$ (negative sign if $x < 0$ ).