Is $x=0$ a local minimum of $f(x)=x^{2/3}(5-x)$? It seems to be decreasing for values below zero and increasing for values between zero and 2.

However $f'(0) = \frac{(-5 (x-2))}{3 x^{1/3}}|_{x=0}$ does not exist.

So it's not a local minimum?

Btw, why does WA seem to say that $x=0$ is not a loc min?



It's not decreasing for values below zero?
I think you made a minor error in taking the derivative. Nonetheless, critical points are points in the domain of $f$ where $f'$ either is zero or fails to exist. So $x=0$ is an ordinary critical point, and a sign analysis for $f'$ shows that a local minimum does indeed occur there. (Determine the signs of the factors of $f'$ and combine them to get the sign of $f'$; signs of factors can change only where they are zero or undefined.)

Incidentally, one shouldn't say "$x=0$ is a local minimum for $f$". $x$ isn't the minimum value, it is the point at which the minimum occurs. The (local) minimum value itself is $f(0)$, which also happens to be $0$.
$$f(x) = x^{2/3}(5 - x)$$

The derivative is

$$f'(x) = x^{2/3}\left(\frac{2}{3}\frac{(5-x)}{x} - 1\right)$$

If we set $f'(x) = 0$ we find two solutions: $x = 0$ which is excluded because it makes the function to explode, and $x = 2$. Indeed if you substitute $x = 2$ you find

$$f'(2) = 2^{2/3}\left(\frac{2}{3}\frac{3}{2} - 1\right) = 0$$

So $x = 2$ is a candidate for being a max or a min. To establish what it is, we study the second derivative. Easy algebra gives

$$f''(x) = -\frac{2}{9}(5-x) x^{-4/3} - \frac{4}{3}x^{-1/3}$$

so

$$f''(2) = -1.32283$$

So we get that $x= 2$ is a \textbf{maximum} point for that function.

Also you can see it in your plot!

Instead, $x = 0$ is not a minimum because the plot doesn't change its curvature. You have like a "bristly jump" as you can see in your plot: the function is not changing from "going upside" to "going downside" or vice versa.