# How do you find the critical points for f(x)= -(sinx)/ (2+cosx)  and the local max and min?

Sep 14, 2016

The critical points are at:
$\left(\frac{2 \pi}{3} , \frac{\sqrt{3}}{3}\right)$is a minimum point

$\left(\left(4 \frac{\pi}{3}\right) , \frac{\sqrt{3}}{3}\right)$ is the maximum point.

#### Explanation:

To find the critical points we have to find $f ' \left(x\right)$
then solve for $f ' \left(x\right) = 0$

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

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

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

Since ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$ we have:
$f ' \left(x\right) = - \frac{2 \cos x + 1}{2 + \cos x} ^ 2$

Let us dolce for $f ' \left(x\right) = 0$to find the critical points:

$f ' \left(x\right) = 0$
$\Rightarrow - \frac{2 \cos x + 1}{2 + \cos x} ^ 2 = 0$
$\Rightarrow - \left(2 \cos x + 1\right) = 0$
$\Rightarrow \left(2 \cos x + 1\right) = 0$
$\Rightarrow 2 \cos x = - 1$
$\Rightarrow \cos x = - \frac{1}{2}$
$\cos \left(\pi - \left(\frac{\pi}{3}\right)\right) = - \frac{1}{2}$

or
$\cos \left(\pi + \left(\frac{\pi}{3}\right)\right) = - \frac{1}{2}$

Therefore,
$x = \pi - \left(\frac{\pi}{3}\right) = \frac{2 \pi}{3}$
or $x = \pi + \left(\frac{\pi}{3}\right) = \frac{4 \pi}{3}$

Let's compute f((2pi)/3)=-sin((2pi)/3)/(2+cos((2pi)/3)

$f \left(\frac{2 \pi}{3}\right) = - \frac{\frac{\sqrt{3}}{2}}{2 - \frac{1}{2}}$
$f \left(\frac{2 \pi}{3}\right) = - \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}$
$f \left(\frac{2 \pi}{3}\right) = - \left(\frac{\sqrt{3}}{3}\right)$
Since$f \left(x\right)$ is decreasing on $\left(0 , \frac{2 \pi}{3}\right)$
Then$\left(\left(\frac{2 \pi}{3}\right) , - \frac{\sqrt{3}}{3}\right)$ is minimum point

Since then the function increases till $x = \left(4 \frac{\pi}{3}\right)$ then the point
$\left(\left(4 \frac{\pi}{3}\right) , \frac{\sqrt{3}}{3}\right)$ is the maximum point.