Find extrema by using the derivative test

Consider $f:[-1,1] \to \mathbb{R}$ defined as $f(x)=x^5$ for each $x\in[-1,1]$. Find maxima and minima of f. Which maxima or minima does the derivative test identify?

Intuitively, maximum is $1$ when $x^*=1$, and minimum is $-1$ when $x^*=-1$. And the candidate by first order condition, $x^*=0$ is a inflection point. But the question is how do I prove these (maximum&minimum) by $n$-th order derivative test?

Since $-1 \le x \le 1 \implies -1 = (-1)^5 \le x^5 \le 1^5 = 1\implies -1 \le f(x) \le 1$, and $f(-1) = -1, f(1) = 1$. So $f_{\text{max}} = 1$, and $f_{\text{min}} = -1$ at $x = 1, -1$ respectively. If you want to use the derivative to solve the problem, you see that $f'(x) = 5x^4 \ge 0$, thus $f$ increases, and you have $-1 = f(-1) \le f(x) \le f(1) = 1$.

First, the question doesn't ask to find maxima and minima using the derivative test. It has two demands:

1. Find maxima and minima of f.
2. Which maxima or minima does the derivative test identify?

Now, the function $f:[-1,1] \to \mathbb{R}, f(x)=x^5$ has only one critical point: $x_0=0$. Because $f'(x)=5x^4$ we see that $f'$ doesn't change the sign in $x_0=0$ therefore $x_0$ is not an extremum point for $f$.
Because $f$ has $2$ extremum points $x_1 = -1, x_2=1$ we conclude that none of them is identified by the derivative test.