# Is f(x)=cos^2x+sin2x increasing or decreasing at x=pi/6?

Increasing at $x = \setminus \frac{\pi}{6}$

#### Explanation:

Given that

$f \left(x\right) = \setminus {\cos}^{2} x + \setminus \sin 2 x$

$f ' \left(x\right) = 2 \setminus \cos x \left(- \setminus \sin x\right) + 2 \setminus \cos 2 x$

$f ' \left(x\right) = - \setminus \sin 2 x + 2 \setminus \cos 2 x$

Now,

$f ' \left(\setminus \frac{\pi}{6}\right) = - \setminus \sin \left(2 \setminus \cdot \setminus \frac{\pi}{6}\right) + 2 \setminus \cos \left(2 \setminus \cdot \setminus \frac{\pi}{6}\right)$

$= - \setminus \sin \left(\setminus \frac{\pi}{3}\right) + 2 \setminus \cos \left(\setminus \frac{\pi}{3}\right)$

$= - \setminus \frac{\sqrt{3}}{2} + 2 \setminus \cdot \frac{1}{2}$

$= \setminus \frac{2 - \setminus \sqrt{3}}{2} > 0$

Since, $f ' \left(\setminus \frac{\pi}{6}\right) > 0$ hence the function $f \left(x\right)$ is increasing at $x = \setminus \frac{\pi}{6}$

Jul 2, 2018

The function is increasing at $x = \frac{\pi}{6}$

#### Explanation:

Calculate the first derivative $f ' \left(x\right)$ and look at the sign of $f ' \left(\frac{\pi}{6}\right)$.

If $f ' \left(\frac{\pi}{6}\right) > 0$, the function is increasing

and

If $f ' \left(\frac{\pi}{6}\right) < 0$, the function is decreasing

The function is

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

Therefore,

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

And

$f ' \left(\frac{\pi}{6}\right) = - \sin \left(2 \cdot \frac{\pi}{6}\right) + 2 \cos \left(2 \cdot \frac{\pi}{6}\right)$

$= - \frac{\sqrt{3}}{2} + 2 \cdot \frac{1}{2}$

$= 1 - \frac{\sqrt{3}}{2}$

$= 0.13$

$f ' \left(\frac{\pi}{6}\right) > 0$

Therefore, the function is increasing at $x = \frac{\pi}{6}$