Question

What are the points of inflection, if any, of `f(x)=x^(1/3)`?

Answer 1

This is an interesting example. The point `(x,f(x))=(0,f(0))=(0,0)` can be considered to be a point of inflection because the graph of `f` changes from concave up to concave down as `x` increases through zero. However, `f'(0)` and `f''(0)` do not exist.

Explanation:

For `f(x)=x^{1/3}`, we have `f'(x)=1/3 x^{-2/3}=1/(3x^{2/3})` when `x !=0` and `f''(x)=-2/9 x^{-5/3}=-2/(9x^{5/3})` when `x !=0`. Therefore, `f''(x) > 0` when `x < 0` and `f''(x) < 0` when `x > 0`.

However, `f'(0)` does not exist since `lim_{h->0}(f(0+h)-f(0))/h=lim_{h->0}(h^{1/3})/h=lim_{h->0}1/(h^{2/3})` does not exist. This also implies that `f''(0)` does not exist.

Even though `f'(0)` does not exist, the graph of `f` does have a vertical tangent line at `x=0` and you can see visually that there is an inflection point there as well in the picture below.

graph{x^(1/3) [-5, 5, -2.5, 2.5]}