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

Question

Christopher Allers · Accepted Answer

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

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]}

Charles Anctil · Answer

undefined To use the first derivative test to determine the local extrema of $ f(x) = x - 2\tan(x) $:

1. Find the first derivative of the function, $ f'(x) $.
2. Solve for critical points by setting $ f'(x) = 0 $ and finding the solutions within the domain of interest.
3. Determine the sign of $ f'(x) $ in the intervals defined by the critical points.
4. Use the sign of $ f'(x) $ to determine the nature of the extrema:
   - If $ f'(x) $ changes sign from positive to negative at a critical point, it indicates a local maximum.
   - If $ f'(x) $ changes sign from negative to positive at a critical point, it indicates a local minimum.
   - If $ f'(x) $ does not change sign at a critical point, the test is inconclusive.

After identifying the nature of the critical points using the first derivative test, you can conclude the presence and type of local extrema for the function $ f(x) = x - 2\tan(x) $.