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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?

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2 Answers 2

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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 $-1 = f(-1) \le f(x) \le f(1) = 1$.

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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?

You did answer to 1. I'll try to help you with an answer for 2.

The first derivative test is based on Fermat theorem

Basically it states that the interior local extremum points are among critical points (points where the derivative is zero). The idea is to find all the critical points and then to check each one to see if it is local extremum.

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.

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