f(x)=x|x|
Show that f'(0) exists. What is the value?
I don't know how to work with the absolute solving for f'(x) with first principles
let's look at it
http://www.wolframalpha.com/input/?i=plot+f(x)%3Dx%7Cx%7C
two cases:
y = x^2 for x≥0 or y = -x^2, for x ≤ 0
remember |x| will always be positive, so the sign can only come from the factor of x at the front
for y = x^2, f '(x) = 2x
f '(0) = 0
for y = -x^2, f '(x) = -2x
f '(0) = 0