How do you find all local maximum and minimum points using the second derivative test given #y=cos^2x-sin^2x#?

1 Answer
Dec 15, 2016

#y(x)# has a local maximum for #x=kπ# and a local minimum for #x=kπ+π/2#

Explanation:

As:

#y=cos^2x-sin^2x=cos(2x)#

we do not really need derivatives because we know that #cos(2x)# has a local maximum for #x=kpi# where #cos(2x)=1# and a local minimum for #x=kpi+pi/2# where #cos(2x)=-1#.

However we can check:

#y'(x) = -2sin(2x)#

so the critical points occur for:

#2x= kpi# or #x=kpi/2#

#y''(x)= -4cos(2x)#

and in fact:

#y''(2kpi) = -4 < 0# so these points are maximums, and

#y''((2k+1pi)) = 4 > 0# so these points are minimums.