Let $$f : R \to R$$ be differentiable at $$c \in R$$ and $$f(c) = 0$$. If $$g(x) = |f(x)|$$, then at $$x = c$$, g is:

We wish to investigate the differentiability of the function $$g(x)=|f(x)|$$ at the point $$x=c$$, given that $$f:\mathbb R \to \mathbb R$$ is differentiable at $$c$$ and $$f(c)=0$$.

Since $$f$$ is differentiable at $$c$$, by definition we have the expansion

$$f(c+h)=f(c)+h\,f'(c)+o(h) \quad\text{as }h\to 0,$$

where $$o(h)$$ denotes a quantity that tends to $$0$$ faster than $$h$$ itself.

Because $$f(c)=0$$, this simplifies to

$$f(c+h)=h\,f'(c)+o(h).$$

Now we examine $$g(c+h)=|f(c+h)|$$. Substituting the above expression, we get

$$g(c+h)=\bigl|\,h\,f'(c)+o(h)\bigr|.$$

We must look at the difference-quotient that defines $$g'(c)$$:

$$\frac{g(c+h)-g(c)}{h} \;=\; \frac{|f(c+h)|-0}{h} \;=\; \frac{\bigl|\,h\,f'(c)+o(h)\bigr|}{h}.$$

Rewrite the numerator by factoring out $$|h|$$:

$$\frac{|h|\,\bigl|f'(c)+o(1)\bigr|}{h} \;=\; \frac{|h|}{h}\;\bigl|f'(c)+o(1)\bigr|,$$

where $$o(1)\to 0$$ as $$h\to 0$$.

Notice that

$$\frac{|h|}{h} = \begin{cases} +1, & h>0,\\[4pt] -1, & h<0. \end{cases}$$

We now distinguish two cases, depending on the value of $$f'(c)$$.

Case 1: $$f'(c)\neq 0$$. Then $$\bigl|f'(c)+o(1)\bigr|\to|f'(c)|\neq 0$$. For $$h>0$$ the quotient tends to $$+|f'(c)|$$, whereas for $$h<0$$ it tends to $$-|f'(c)|$$. Since the right-hand and left-hand limits are unequal, the limit does not exist. Hence $$g$$ is not differentiable at $$x=c$$ when $$f'(c)\neq 0$$.

Case 2: $$f'(c)=0$$. Our expansion becomes $$f(c+h)=o(h)$$, so

$$\bigl|f(c+h)\bigr|=|o(h)|=o(h).$$

Therefore

$$\frac{|f(c+h)|}{h}=\frac{o(h)}{h}=o(1)\to 0\quad\text{as }h\to 0.$$

The limit from both sides equals $$0$$, so $$g'(c)=0$$ exists. Thus $$g$$ is differentiable at $$x=c$$ when $$f'(c)=0$$.

Combining the two cases, we conclude that $$g$$ is differentiable at $$x=c$$ if and only if $$f'(c)=0$$.

Hence, the correct answer is Option C.