Analyzing a function with cube roots

Question

I am analyzing this function: \begin{align} f(x) = \sqrt[3]{ x } - \sqrt[3]{x+1} \end{align}

I proved it has a horizontal asymptote at y=0. However, on the graph, it has a minimum for x=-0.5 but I can't find that. First derivative has no zeroes, so the function has no stacionary points. Second derivative also has no zeroes. How do I find that minimum?

Answer 1

You must've taken the first derivative wrong:

$$f'(x)=\frac13(x^{-2/3}-(x+1)^{-2/3})$$

Setting $f'(x)=0$,

$$0=x^{-2/3}-(x+1)^{-2/3}$$

$$x^{-2/3}=(x+1)^{-2/3}$$

$$\pm x=x+1$$

$$x=-1/2$$

Answer 2

Don't forget that critical points also occur where the derivative is undefined, as long as the point in question is in the domain of the original function.

Since $f(x) = x^{1/3} - (x+1)^{1/3}$, then $f'(x) = \dfrac13 x^{-2/3} - \dfrac13 (x+1)^{-2/3}$.

Note that $f'(x)$ is undefined at $x=0$ and $x=-1$, and both of these are in the domain of $f(x)$, so there may be extrema at those locations.