Question

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

Answer 1

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.