# How do you find the absolute minimum and maximum on [-pi/2,pi/2] of the function f(x)=sinx^2?

• If $x$ runs on $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, then ${x}^{2}$ runs on $\left[0 , {\pi}^{2} / 4\right]$. So, you are looking for min and max of $\sin$ on the interval $\left[0 , {\pi}^{2} / 4\right]$.
• Because $\frac{\pi}{2} \in \left[0 , {\pi}^{2} / 4\right]$, the maximum is $1 = \sin \left(\frac{\pi}{2}\right)$.
• Because ${\pi}^{2} / 4 < \pi$, sin is positive on $\left[0 , {\pi}^{2} / 4\right]$, so the minimum is $0 = \sin \left(0\right)$.