How do you find the critical numbers of #y= 2x-tanx#?

1 Answer
Nov 5, 2016

Answer:

#x=pi/4+(kpi)/2,pi/2+kpi,kinZZ#

Explanation:

Note that a critical number of a function #f# will occur at #x=a# when #f'(a)=0# or #f'(a)# is undefined.

So, we first need to find the derivative of the function.

#y=2x-tanx#

#dy/dx=2-sec^2x#

So, critical values will occur when:

#0=2-sec^2x#

Or:

#sec^2x=2" "=>" "secx=+-sqrt2" "=>" "cosx=+-1/sqrt2=+-sqrt2/2#

Note that this occurs at #x=pi/4,(3pi)/4,(5pi)/4,(7pi)/4...# which can be summarized as #x=pi/4+(kpi)/2# where #kinZZ#, which means that #k# is an integer.

Furthermore, note that #2-sec^2x# is undefined for some values:

#2-sec^2x=2-1/cos^2x#

This is undefined when #cosx=0#, which occurs at #x=pi/2,(3pi)/2,(5pi)/2...# or #x=pi/2+kpi,kinZZ#.

So, we have critical values at #x=pi/4+(kpi)/2,pi/2+kpi,kinZZ#.