# Question #56aac

Oct 2, 2016

Local minimum is 1 at $x = 2 k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$
Local maximum is -1 at $x = \left(2 k + 1\right) \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$
There is no global maximum nor minimm.

#### Explanation:

Here, $y = \sec x = \frac{1}{\cos} x$.

For $x \in \left(0 , \frac{\pi}{2}\right) , y \in \left(1 , \infty\right)$

For $x \in \left(\frac{\pi}{2} , \pi\right] , y \in \left(- \infty , - 1\right)$

For $x \in \left[\pi , \frac{3}{2} \pi\right) , y \in \left(- 1 , - \infty\right)$

For $x \in \left(\frac{3}{2} \pi , 2 \pi\right) , y \in \left(\infty , 1\right)$

The period of sec x being $2 \pi$, the ranges repeat this way,

for $x \in \left(2 k \pi , 2 \left(k + 2\right) \pi\right) , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

Note the infinite discontinuities at $x = \frac{\pi}{2} \mathmr{and} \frac{3}{2} \pi$

Local minimum is 1 at $x = 2 k \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

Local maximum is -1 at $x = \left(2 k + 1\right) \pi , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

There is no global maximum nor minimm