Is $x=0$ a local minimum of $f(x)=x^{2/3}(5-x)$?

Question

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?

Answer 1

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 $0$.

Answer 2

Near $x=0$, $f(x) \sim 5 x^{2/3}$. Hence $f(0)=0$ and $f(x) \geq 0$ in a neighborhood of $x=0$. This means that $x=0$ is a local minimum.

I would like to remark that $x^{2/3}=\sqrt[3]{x^2}=\left( \sqrt[3]{x} \right)^2$, and therefore it can be computed on the whole real axis. Wolfram Alpha probably assumes that $$x^{2/3} = \exp \left( \frac23 \log x \right),$$ but this representation is valid for $x >0$ only.

The fact that the derivative does not exist is not an abstruction: think of $x \mapsto |x|$. If you are thinking of the Fermat Theorem, it precisely states that if a function attains a local minimum (or maximum) at $x_0$ and the function is differentiable at $x_0$, then the derivative at $x_0$ vanishes. In your case, the function is not differentiable at the minimum point, so that the theorem does not apply.

Answer 3

$$f(x) = x^{2/3}(5 - x)$$

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.

Answer 4

The given function $f$ is not differentiable at $x=0$, let alone twice differentiable. As a consequence the usual test for a local minimum cannot be applied here. But we can argue as follows:

Since $f(0)=0$ and $f(x)>0$ for $0<|x|<1$ we have a local minimum at $x=0$.