# How do you find the critical points for f(x)=sinx cosx  and the local max and min in #0 ≤ x < 2pi?

Jan 24, 2017

$x = \frac{\pi}{4}$: Local max
$x = \frac{3 \pi}{4}$: Local Min

#### Explanation:

Differentiate:

$f ' \left(x\right) = \cos x \left(\cos x\right) + \sin x \left(- \sin x\right)$

$f ' \left(x\right) = {\cos}^{2} x - {\sin}^{2} x$

$f ' \left(x\right) = \cos 2 x$

Critical points will occur when the derivative equals $0$ or is undefined.

There will be no undefined points on the function, since it's continuous.

$0 = \cos 2 x$

$2 x = \frac{\pi}{2} \mathmr{and} \frac{3 \pi}{2}$

$x = \frac{\pi}{4} \mathmr{and} \frac{3 \pi}{4}$

Now check the signs to the left and right of each of these points.

$f ' \left(0\right) = \cos \left(2 \left(0\right)\right) = \cos \left(0\right) = 1$

$f ' \left(\frac{\pi}{2}\right) = \cos \left(2 \left(\frac{\pi}{2}\right)\right) = \cos \left(\pi\right) = - 1$

So $x = \frac{\pi}{4}$ is a local maximum (the derivative is increasing, then decreasing).

$f ' \left(\frac{2 \pi}{3}\right) = \cos \left(2 \left(\frac{2 \pi}{3}\right)\right) = \cos \left(\frac{4 \pi}{3}\right) = - \frac{1}{2}$

$f ' \left(\frac{5 \pi}{6}\right) = \cos \left(2 \left(\frac{5 \pi}{6}\right)\right) = \cos \left(\frac{5 \pi}{3}\right) = \frac{1}{2}$

So, $x = \frac{3 \pi}{4}$ is a local minimum.

Hopefully this helps!