Let #f(x)= sqrt(|x^3+1|)#. How to show that #f(x)# is not differentiable at #x=-1#?

1 Answer
Oct 20, 2017

Please see below.

Explanation:

Quick and maybe sketchy

#f(x) = {(sqrt(x^3+1), x >= -1),(sqrt(-x^3-1),x <= -1):}#

The left and right derivatives at #-1# do not exist.
(Specifically, the right derivative increases without bound and the left decreases without bound as #xrarr-1#.)

#f'(x) = {((3x^2)/(2sqrt(x^3+1)), x > -1),((-3x^2)/(2sqrt(-x^3-1)),x < -1):}#

More detailed
Using the definition of #f# above (piecewise and without absolute value),

Look at the limit from the right:

#lim_(xrarr-1^+)(f(x)-f(-1))/(x+1)# or #lim_(hrarr0+)(f(-1+h)-f(-1))/h#

or the limit from the left:

#lim_(xrarr-1^-)(f(x)-f(-1))/(x+1)# or #lim_(hrarr0-)(f(-1+h)-f(-1))/h#

to see that the two-sided limit cannot exist.