Range of $m$ for $f(x)=\sin^3x-m\sin x$ to have one maximum and one minimum on $(-\pi/2,\pi/2)$ The function $f(x)=\sin^3x-m\sin x$ is defined on $(-\pi/2,\pi/2)$ and it assumes only one maximum value and only one minimum value on this interval. Then the range for $m$ is: …?
My attempt:
$$f'(x)=3\sin^2x\cos x-m\cos x$$
$$=\cos x(3\sin^2x-m)=0$$
So, $\cos x=0$ or $\sin^2x=m/3$.
$$\cos x=0$$ is not possible since $x\in(-\pi/2,\pi/2)$.
$$f''(x)=6\sin x\cos^2 x-3\sin^3 x+m\sin x$$
I'm stuck after this.
The solution is given as $0<m<3$.
 A: Hint:
At the minimum and maximum, $\sin^2x=m/3$, so $0\le m/3\le 1$. Study separately the cases $m=0$ and $m=3$.
A: First of all, your second derivative is incorrect.
Now, since the endpoints aren't included with the interval, the points of maximum and minimum would have to be points of local maximum and minimum in the interior of the interval. Thus, they have to be critical points satisfying $f'(x)=0$. You've correctly reduced this equation to
$$\sin^2x=m/3.$$
Note that this equation wouldn't have any solutions if $m<0$ or if $m>3$, so it has to be in between $0\le m\le3$.


*

*If $m=0$, then $\sin^2x=0$ gives us only one critical point $x=0$ (in the given interval, of course). Regardless of whether it's a local maximum or minimum, it's not good enough, because we need both — so we need two critical points, not one.

*If $m=3$, then $\sin x=\pm1$, which has no solutions and thus no critical points in the given interval.


So we narrowed it down to $0<m<3$. For any such $m$:
$$\sin^2x=\frac{m}{3} \implies \sin x=\pm\sqrt{\frac{m}{3}} \implies x=\pm\arcsin\left(\sqrt{\frac{m}{3}}\right),$$
precisely two solutions lying within $(-\pi/2,\pi/2)$. And it shouldn't be difficult to check, either using the first derivative test or the second derivative test (with the correct second derivative) that one of them is a maximum and one is a minimum.
