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

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Answer 1 · 2017-02-24T06:47:31

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]}

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