How to show f(x)=|x|f(x)=|x| It is differentiable everywhere except at the point x = 0 ?

1 Answer
MattyMatty
Feb 9, 2018

"See explanation"See explanation

Explanation:

"Apply the definition of |x| : "Apply the definition of |x| :
f(x) = |x| => f(x)=|x|⇒
{ ( f(x) = x, x >= 0 ), ( f(x) = -x, x<= 0 ) :}
"Now derive : "
{ ( f '(x) = 1, x >= 0), ( f '(x) = -1, x <= 0 ) :}
"So we see there is a discontinuity in x=0 for f '(x)."
"For the rest, it is differentiable everywhere."

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