# How do you find the local extrema for f(x) = 3x^5 - 10x^3 - 1 on the interval [-1,1]?

Apr 7, 2016

That function has no local extrema on that interval. (It does, of course, have global extrema on the closed interval.)

#### Explanation:

I take the definition of local maximum to be:

$f \left(c\right)$ is a local maximum for $f$ if and only if the is an open interval, $I$, containing $c$ such that $f \left(c\right) \ge f \left(x\right)$ for all $x \in I$

(For local minimum, reverse the inequality.)

Fermat's Theorem tells us that $f \left(c\right)$ is a local extremum if and only if $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist.

For $f \left(x\right) = 3 {x}^{5} - 10 {x}^{3} - 1$, we have

$f ' \left(x\right) = 15 {x}^{4} - 30 {x}^{2} = 15 {x}^{2} \left({x}^{2} - 2\right)$.

$f '$ never fails to exist and it is $0$ at $x = 0$, $- \sqrt{2}$ and $\sqrt{2}$. But the only one of the in the interval is $x = 0$.

Applying the first derivative test, we see that $f \left(0\right)$ is neither a minimum nor a maximum. (Near $0$ on either side, we have $f ' \left(x\right) < 0$, so $f$ is decreasing on both sides of $0$.)

Alternatives

Some presentations of calculus may include clauses in the definition so that we need only an open interval, $I$, such that for all $x$ in the intersection of $I$ and the domain of $f$, we have $f \left(c\right) \ge x$
In such a presentation, since we have restricted the domain of $f$ to $\left[- 1 , 1\right]$, we must answer that $f \left(1\right) = - 8$ in a local minimum and $f \left(- 1\right) = 6$ is a local maximum.