Let f:RR be differentiable at cR and f(c)=0.

Question

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

Accepted Answer

$$g'(c) = \mathop {\lim }\limits_{x \to c} {{g(x) - g(c)} \over {x - c}}$$

$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right| - \left| {f(c)} \right|} \over {x - c}}$$

$$ \therefore $$ f(c) = 0

$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}$$

$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}}$$ if f(x) > 0

and $$g'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}}$$ if f(x) < 0

$$ \Rightarrow g'(c) = f'(c) =  - f'(c)$$

$$ \Rightarrow $$ 2f'(c) = 0

$$ \Rightarrow $$ f'(c) = 0

Solutions · 1