Question

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

Answer 1

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

Explanation:

If `f(x)=x-2tan(x)` then `f'(x)=1-2sec^2(x)`. Setting this equal to zero results in the equation `sec^2(x)=1/2`, which is equivalent to `cos^2(x)=2`, which clearly has no solutions.

In fact, `sec^2(x) geq 1` for all `x` where it is defined so that `f'(x)=1-2sec^2(x) leq 1-2=-1` for all `x` where `f(x)` is defined, so `f(x)` is strictly decreasing over individual intervals on which it is defined (such as the interval `(-pi/2,pi/2)`).

Here's the graph of `f`:

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