How do you find the critical numbers of #y = sin^2 x#?

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Answer 1 · 2016-03-03T23:13:23

We know that #cos2x=cos^2x-sin^2x=>cos2x=1-2*sin^2x=> sin^2x=1/2[1-cos2x]#

Hence #y# becomes #y=1/2*[1-cos2x]#

Now the critical values #c# of #f# are those if and only if

i. #f'(c)=0#
ii. #f'(c)# is undefined

So the critical values are those for which #dy/(dc)=0# hence

#dy/(dc)=sin2c=0=>2c=n*pi=>c=n*pi/2#

where n is an integer.

The critical points are at #x=(n*pi)/2#