# How do use the first derivative test to determine the local extrema f(x)=x-2tan(x)?

Sep 15, 2015

This function has no local extreme points because its derivative is never zero.

#### Explanation:

If $f \left(x\right) = x - 2 \tan \left(x\right)$ then $f ' \left(x\right) = 1 - 2 {\sec}^{2} \left(x\right)$. Setting this equal to zero results in the equation ${\sec}^{2} \left(x\right) = \frac{1}{2}$, which is equivalent to ${\cos}^{2} \left(x\right) = 2$, which clearly has no solutions.

In fact, ${\sec}^{2} \left(x\right) \ge q 1$ for all $x$ where it is defined so that $f ' \left(x\right) = 1 - 2 {\sec}^{2} \left(x\right) \le q 1 - 2 = - 1$ for all $x$ where $f \left(x\right)$ is defined, so $f \left(x\right)$ is strictly decreasing over individual intervals on which it is defined (such as the interval $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$).

Here's the graph of $f$:

graph{x-2tan(x) [-20, 20, -10, 10]}