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

1 Answer
Feb 24, 2017

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#

Explanation:

#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]}