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

Dec 1, 2016

$f \left(x\right)$ has a maximum in $x = - \frac{\pi}{3} + 2 k \pi$ and a minimum in $x = \frac{\pi}{3} + 2 x \pi$

#### Explanation:

Critical points are where the first derivative is zero.

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

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

So critical points are where:

$1 - 2 \cos x = 0$

$\cos x = \frac{1}{2}$

$x = \frac{\pi}{3} + 2 k \pi$ and $x = - \frac{\pi}{3} + 2 k \pi$

Without calculating the second derivative, we can see that the denominator of $f ' \left(x\right)$ is always positive, so the sign of the derivative is the sign of $1 - 2 \cos x$.

So, for ${x}_{k} = \frac{\pi}{3} + 2 k \pi$, $f ' \left({x}_{k} - \epsilon\right) < 0$ and $f ' \left({x}_{k} + \epsilon\right) > 0$ so these points are minimums, while for ${x}_{k} = - - \frac{\pi}{3} + 2 \pi$ the opposite is true and these points are maximums.

graph{sinx/(cosx-2) [-10, 10, -5, 5]}