Extrema of $f(x)= x^2e^{\sin(x)}$

Question

I was given the task show that $f(x):= x^2e^{\sin(x)}$ defined on $[-\frac{\pi}{2},\frac{\pi}{2}]$ has three extrema.

I also have to determine of which kind they are (minimum/ maximum).

I plotted the graph of $f$ and this confused me since according to the graph, $f$ has just $2$ extrema for $x \in[-\frac{\pi}{2},\frac{\pi}{2}]$.

Answer 1

1. Compute $f(\pi/2)$ and $f(- \pi/2)$.

2. Show that for $x \in [- \pi/2, \pi/2]$ we have $f'(x)=0 \iff x=0.$

3. Show that for $x \in [- \pi/2, \pi/2]$ we have $f'(x)<0 \iff x=[ - \pi/2,0)$

4. Show that for $x \in [- \pi/2, \pi/2]$ we have $f'(x)>0 \iff x=(0, \pi/2]$.

The consequences are: $f$ has local extrema exactly for $- \pi/2,0$ and $ \pi/2.$

Answer 2

We wish to find the relative and absolute extrema of the function $f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \to \mathbb{R}$ defined by $$f(x) = x^2e^{\sin x}$$ The extrema can occur at a critical point, a point in the interior of the interval where the function is not defined, or at an endpoint. $$f'(x) = 2xe^{\sin x} + x^2e^{\sin x}\cos x$$ which is defined everywhere in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

To find the critical points, we set the derivative equal to zero. \begin{align*} f'(x) & = 0\\ 2xe^{\sin x} + x^2e^{\sin x}\cos x & = 0\\ xe^{\sin x}(2 + x\cos x) & = 0 \end{align*} By inspection, $f'(x) = 0$ if $x = 0$. Are there any other critical points?

If $x \in \left(0, \frac{\pi}{2}\right)$, then $x > 0$, $e^{\sin x} > 0$, $\cos x > 0$, so the factors $xe^{\sin x}$ and $2 + x\cos x > 0$. Thus, $f'(x) > 0$. Hence, there are no critical points in the interval $\left(0, \frac{\pi}{2}\right)$.

If $x \in \left(-\frac{\pi}{2}, 0\right)$, then $x < 0$ and $e^{\sin x} > 0$, so the factor $xe^{\sin x} < 0$. If $-\frac{\pi}{2} < x < 0$, then $0 < \cos x < 1$ and $x < 0$, so $x < x\cos x$. Thus, $2 + x\cos x > 2 + x > 2 - \frac{\pi}{2} > 0$. Therefore, $f'(x) < 0$ on the interval $\left(-\frac{\pi}{2}, 0\right)$, so there are no critical points in this interval.

Hence, $0$ is the only critical point. By the First Derivative Test, since $f'(x)$ changes from negative to positive at $x = 0$, $x = 0$ is a relative minimum. The relative minimum value is $f(0) = 0$.

It is possible for the function $f$ to have absolute extrema occur at the endpoints.
\begin{align*} f\left(-\frac{\pi}{2}\right) & = \left(-\frac{\pi}{2}\right)^2e^{\sin\left(-\frac{\pi}{2}\right)} = \frac{\pi^2}{4e}\\ f\left(\frac{\pi}{2}\right) & = \left(\frac{\pi}{2}\right)^2e^{\sin\left(\frac{\pi}{2}\right)} = \frac{\pi^2e}{4} \end{align*} Both of these values are positive, so the relative minimum is also an absolute minimum. Thus, the function has an absolute minimum value of $0$ at $x = 0$. Since $f\left(\frac{\pi}{2}\right) > f\left(-\frac{\pi}{2}\right)$, the absolute maximum value of $\frac{\pi^2e}{4}$ occurs at $x = \frac{\pi}{2}$.

The function has only two extrema on this interval, as you observed by examining its graph.