Question

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

Answer 1

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]}