# How do you find f'(x) using the definition of a derivative for f(x)=absx?

Oct 17, 2015

See the explanation.

#### Explanation:

The function $f \left(x\right) = | x |$ is continuous function, defined for $\forall x \in R$ but it's not differentiable for $\forall x \in R$ (e.g. derivative doesn't exist in every point).

By definition, function is differentiable at a point if its derivative exists at that point.

$f \left(x\right) = x$ for $x \ge 0$
$f \left(x\right) = - x$ for $x < 0$

For $x > 0$:
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{x + h - x}{h} = {\lim}_{h \to 0} \frac{h}{h} = {\lim}_{h \to 0} 1 = 1$

For $x < 0$:
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{- \left(x + h\right) - \left(- x\right)}{h} = {\lim}_{h \to 0} \frac{- x - h + x}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{- h}{h} = {\lim}_{h \to 0} \left(- 1\right) = - 1$

From the previous:

$f ' \left(0 + \epsilon\right) = 1$
$f ' \left(0 - \epsilon\right) = - 1$

$f ' \left(0 + \epsilon\right) \ne f ' \left(0 - \epsilon\right)$

Function is not differentiable at $x = 0$.