# How do you find the critical numbers of y= 2x-tanx?

Nov 5, 2016

$x = \frac{\pi}{4} + \frac{k \pi}{2} , \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$

#### Explanation:

Note that a critical number of a function $f$ will occur at $x = a$ when $f ' \left(a\right) = 0$ or $f ' \left(a\right)$ is undefined.

So, we first need to find the derivative of the function.

$y = 2 x - \tan x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 - {\sec}^{2} x$

So, critical values will occur when:

$0 = 2 - {\sec}^{2} x$

Or:

${\sec}^{2} x = 2 \text{ "=>" "secx=+-sqrt2" "=>" } \cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

Note that this occurs at $x = \frac{\pi}{4} , \frac{3 \pi}{4} , \frac{5 \pi}{4} , \frac{7 \pi}{4.} . .$ which can be summarized as $x = \frac{\pi}{4} + \frac{k \pi}{2}$ where $k \in \mathbb{Z}$, which means that $k$ is an integer.

Furthermore, note that $2 - {\sec}^{2} x$ is undefined for some values:

$2 - {\sec}^{2} x = 2 - \frac{1}{\cos} ^ 2 x$

This is undefined when $\cos x = 0$, which occurs at $x = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2.} . .$ or $x = \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$.

So, we have critical values at $x = \frac{\pi}{4} + \frac{k \pi}{2} , \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$.