Let f(x) be a function satisfying |f(x)| ≤ x^2 for -1 ≤ x ≤ 1, how do you show that f is differentiable at x = 0 and find f’(0)?

1 Answer
Apr 20, 2018

The derivative exists and is zero.

Explanation:

Since #|f(x)|<=x^2# for #xin [-1,1]#, we must have
#|f(0)| <= 0^2=0#, but since it is definitely non-negative, it must be 0.

#f(0) = 0#

Now

#f^'(0) = lim_{h to 0} (f(0+h)-f(0))/h = lim_(h to 0) f(h)/h#

Thus

#|f^'(0)| = lim_{h to 0}|f(h)/h|= lim_{h to 0} |f(h)|/h #
#qquad <= lim_(h to 0) h^2/h=lim_(h to 0) h = 0#

Hence #|f^'(0)| <= 0 implies |f^'(0)| = 0#

Hence the derivative at #x=0# exists and is 0.