# Determine where the function isn't differentiable and graph f(x)=|x|

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By definition,

`f'(a)=lim_(h->0) (|a+h|-|a|)/h` whenever this limit exists.

If `a>0,` then for small enough values of `h,` `a+h>0` as well. By definition of absolute value, `|x|=x` if `x>=0.` Thus if `a>0,`

`f'(a)=lim_(h->0) (|a+h|-|a|)/h=lim_(h->0) (a+h-a)/h=lim_(h->0)h/h=1.`

If `a<0,` then for small enough values of `h,` `a+h<0` as well. By definition of absolute value, `|x|=-x` if `x<0.` Thus if `a<0,`

`f'(a)=lim_(h->0) (|a+h|-|a|)/h=lim_(h->0) (-a-h+a)/h=lim_(h->0)(-h)/h=-1.`

When `a=0,` we have

`lim_(h->0)(|0+h|-|0|)/h=lim_(h->0)(|h|)/h.`

If `h<0,` `|h|/h=-h/h=-1.` If `h>0,` `|h|/h=h/h=1.` Therefore this limit doesn't exist.

**We've proved that** `f`** is differentiable for all** `x` **except** `x=0.` It can be proved that if a function is differentiable at a point, then it is continuous there. The absolute value function shows that the converse is not true. Here's the graph.

**Sources:**