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

1 Answer
Sep 15, 2015

Answer:

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]}