Let f : R to R be differentiable at c in R an | vedclass

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(A) We examine the differentiability of $g(x) = |f(x)|$ at $x = c$ using the definition of the derivative:$g'(c) = \lim_{h 	o 0} \frac{|f(c + h)| - |

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differentiable if $f'(c) = 0$

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(A) We examine the differentiability of $g(x) = |f(x)|$ at $x = c$ using the definition of the derivative:$g'(c) = \lim_{h 	o 0} \frac{|f(c + h)| - |f(c)|}{h}$Since $f(c) = 0$,this simplifies to:$g'(c) = \lim_{h 	o 0} \frac{|f(c + h)|}{h}$Using the definition of the derivative $f'(c) = \lim_{h 	o 0} \frac{f(c + h) - f(c)}{h} = \lim_{h 	o 0} \frac{f(c + h)}{h}$,we can write:$g'(c) = \lim_{h 	o 0} \left| \frac{f(c + h)}{h} ight| \cdot \frac{|h|}{h} = |f'(c)| \cdot \lim_{h 	o 0} \frac{|h|}{h}$If $f'(c) = 0$,then $g'(c) = 0 \cdot (\pm 1) = 0$,which means $g(x)$ is differentiable at $x = c$.If $f'(c) 
eq 0$,the limit $\lim_{h 	o 0} \frac{|h|}{h}$ does not exist (it is $1$ for $h > 0$ and $-1$ for $h Therefore,$g(x)$ is differentiable at $x = c$ if $f'(c) = 0$.

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

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