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

Nov 10, 2016

graph{(tanx)^2 [-7.09, 12.91, -2.12, 7.88]}

It has minimum at $x = k \pi , k \in \mathbb{Z}$ valued 0

#### Explanation:

$y ' = 2 \frac{\tan x}{\cos} ^ 2 x$

$y ' = 0 , \forall x = k \pi , k \in \mathbb{Z}$

$y ' ' = 2 \cdot \frac{\frac{1}{\cos} ^ 2 x \cdot {\cos}^{2} x + 2 \tan x \cos x \sin x}{{\cos}^{4} x} = 2 \cdot \frac{1 + 2 {\sin}^{2} x}{\cos} ^ 4 x > 0 , \forall x \in \mathbb{R}$

So the critical points in $k \pi$ are local minimum.