# How do you find all local maximum and minimum points using the second derivative test given y=cos^2x-sin^2x?

Dec 15, 2016

$y \left(x\right)$ has a local maximum for x=kπ and a local minimum for x=kπ+π/2

#### Explanation:

As:

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

we do not really need derivatives because we know that $\cos \left(2 x\right)$ has a local maximum for $x = k \pi$ where $\cos \left(2 x\right) = 1$ and a local minimum for $x = k \pi + \frac{\pi}{2}$ where $\cos \left(2 x\right) = - 1$.

However we can check:

$y ' \left(x\right) = - 2 \sin \left(2 x\right)$

so the critical points occur for:

$2 x = k \pi$ or $x = k \frac{\pi}{2}$

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

and in fact:

$y ' ' \left(2 k \pi\right) = - 4 < 0$ so these points are maximums, and

$y ' ' \left(\left(2 k + 1 \pi\right)\right) = 4 > 0$ so these points are minimums.