Find the maximum and minimum points of $f(x)=x^{\frac{2}{3}}$ Find the maximum and minimum points of $f(x)=x^{\frac{2}{3}}$ on the interval $[-1,1]$
Do these occur at critical points?
So I have the first derivative as
$f'(x) = \frac{2}{3}x^{-\frac{1}{3}}$ and then when I set it equal to zero there are no solutions so where do I go from here?
 A: This function is not even differentiable in the point $x=0$ since $$\lim_{x\to 0^+}f'(x)=+\infty$$ and $$\lim_{x\to 0^+}f'(x)=-\infty$$ So the criterion of the first derivative for a max/min does not apply! 
Split the interval in two intervals $I_1=[-1,0)$ and $I_2=(0,1]$. In these intervals there is no problem with the derivative. In $I_1$ $f$ is monotone decreasing and in $I_2$ monotone increasing. So the only candidates for extrema are the points $\{-1,1\}$ for maxima and $\{0\}$ for minimum (but not with the use of the derivative criterion, since it does not apply!) Find the solutions by checking the candidates (there are only three of them).
A: $$f'(x) = \frac{2}{3}x^{-\frac{1}{3}}$$
The question is asking you about the maximum and minimum points of $f(x)$ in that specific interval. Setting the derivative to $0$ will give you the absolute minima/maxima, which are not necessarily the maximum/minimum points in the interval.
$$0 = \frac{2}{3}x^{-\frac{1}{3}}$$
This function has no absolute minima/maxima; the derivative will never equal $0$. It makes sense from the graph too. However, since you're looking for the minimum and maximum points in the interval, you're just looking for places where $f(x)$ is lowest or highest between $-1$ and $1$ (inclusive).

By graphing, we see that it is lowest at $(0, 0)$, making it a minimum, and highest at $(\pm1,1)$, the relative maximum. Again, these are not critical points, but they are your maxima/minima.
Without graphing, the only real way you can do this is by noticing that it's $f(x)=x^{\frac{2}{3}}=(x^2)^{\frac{1}{3}}$. In other words, negative or positive $x$, $y$ will always be positive. Also that $y$ is increasing as $|x|$ increases. Meaning that 0 is the lowest, and $\pm1$ is highest in the interval.
