How do you find the critical points for #f(x)=3sin^2 x# and the local max and min?

Question

Charles Anctil · Accepted Answer

Critical points:
#f(x)_max = 3# for #x=((2n-1)pi)/2 forall n in ZZ#
#f(x)_min = 0# for #x=(npi) forall n in ZZ#

Local max: #3#, Local min: #0# #f(x) = 3sin^2x# #f'(x) = 3* 2sinx*cosx# [Chain rule] #= 3sin2x# For critical points #f'(x) = 0# #:. 3sin2x = 0# will yield critical points #(x, f(x))# #sin 2x = 0 -> 2x = 0, pi, 2pi, 3pi# for #2x in [0, 3pi]# #:. x=0, pi/2, pi, (3pi)/2, .....# and #= -pi/2, -pi, -(3pi)/2, ....# In general: #f'(x)=0# for #x= (npi)/2 forall n in ZZ# #f(x)_max = 3*(+-1)^2= 3# #f(x)_min = 3*0 =0# Hence critical points are: #f(x)_max = 3# for #x=((2n-1)pi)/2 forall n in ZZ# and #f(x)_min = 0# for #x=(npi) forall n in ZZ# #:. # Local max: #3#, Local min: #0# These points can be seen on the graph of #f(x)# below: graph{3*(sinx)^2 [-10, 10, -5, 5]}

Christopher Allers · Answer

undefined To find the critical points of $ f(x) = 3\sin^2(x) $, first, take the derivative of $ f(x) $ with respect to $ x $ and set it equal to zero. Then solve for $ x $. After finding the critical points, you can determine whether they correspond to local maxima, minima, or neither by using the first or second derivative test.