# What are the critical values, if any, of f(x)=2secx + tan x in[-pi,pi]?

Nov 26, 2016

$x = - \frac{5 \pi}{6} , - \frac{\pi}{2} , - \frac{\pi}{6} , \frac{\pi}{2}$

#### Explanation:

Quickly showing the derivations for the derivatives of $\sec x = \frac{1}{\cos} x$ and $\tan x = \sin \frac{x}{\cos} x$:

$\frac{d}{\mathrm{dx}} \sec x = \frac{d}{\mathrm{dx}} \left(\frac{1}{\cos} x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sec x} = \frac{d}{\mathrm{dx}} {\left(\cos x\right)}^{-} 1$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sec x} = - {\left(\cos x\right)}^{-} 2 \frac{d}{\mathrm{dx}} \left(\cos x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sec x} = - \frac{1}{\cos} ^ 2 x \left(- \sin x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sec x} = \frac{1}{\cos} x \left(\sin \frac{x}{\cos} x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \sec x} = \sec x \tan x$

$\frac{d}{\mathrm{dx}} \tan x = \frac{d}{\mathrm{dx}} \left(\sin \frac{x}{\cos} x\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \tan x} = \frac{\left(\frac{d}{\mathrm{dx}} \sin x\right) \cos x - \sin x \left(\frac{d}{\mathrm{dx}} \cos x\right)}{\cos} ^ 2 x$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \tan x} = \frac{\cos x \left(\cos x\right) - \sin x \left(- \sin x\right)}{\cos} ^ 2 x$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \tan x} = \frac{{\cos}^{2} x + {\sin}^{2} x}{\cos} ^ 2 x$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \tan x} = \frac{1}{\cos} ^ 2 x$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \tan x} = {\sec}^{2} x$

So, if $f \left(x\right) = 2 \sec x + \tan x$, then:

$f ' \left(x\right) = 2 \sec x \tan x + {\sec}^{2} x$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{2 \sin x}{\cos} ^ 2 x + \frac{1}{\cos} ^ 2 x$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{2 \sin x + 1}{\cos} ^ 2 x$

A critical value will occur when $f ' \left(x\right) = 0$ or when $f '$ is undefined.

We note that $f ' \left(x\right) = 0$ when $2 \sin x + 1 = 0$. This means that $\sin x = - \frac{1}{2}$. On the interval $x \in \left[- \pi , \pi\right]$ this means there are critical values at color(blue)(x=-pi/6,-(5pi)/6.

$f '$ is undefined when ${\cos}^{2} x = 0$, so $\cos x = 0$. In the interval this is at color(blue)(x=-pi/2,pi/2.