# What are the extrema of g(x) = 2 sin(2x - pi) + 4 on [-pi/2,pi/2]?

May 27, 2018

$x = \pm \frac{\pi}{4}$ for $x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

#### Explanation:

$g \left(x\right) = 2 \sin \left(2 x - \pi\right) + 4$

$g \left(x\right) = - 2 \sin \left(2 x\right) + 4$

For extrema of $g \left(x\right)$, $g ' \left(x\right) = 0$

$g ' \left(x\right) = - 4 \cos \left(2 x\right)$

$g ' \left(x\right) = 0$

$- 4 \cos \left(2 x\right) = 0$

$\cos \left(2 x\right) = 0$

$2 x = \pm \frac{\pi}{2}$

$x = \pm \frac{\pi}{4}$ for $x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$