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

If you are being graded based on your answer, check you grader's definitions of local extrema.

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.