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

Feb 24, 2017

Critical points:
$f {\left(x\right)}_{\max} = 3$ for $x = \frac{\left(2 n - 1\right) \pi}{2} \forall n \in \mathbb{Z}$
$f {\left(x\right)}_{\min} = 0$ for $x = \left(n \pi\right) \forall n \in \mathbb{Z}$

Local max: $3$, Local min: $0$

#### Explanation:

$f \left(x\right) = 3 {\sin}^{2} x$

$f ' \left(x\right) = 3 \cdot 2 \sin x \cdot \cos x$ [Chain rule]

$= 3 \sin 2 x$

For critical points $f ' \left(x\right) = 0$

$\therefore 3 \sin 2 x = 0$ will yield critical points $\left(x , f \left(x\right)\right)$

$\sin 2 x = 0 \to 2 x = 0 , \pi , 2 \pi , 3 \pi$ for $2 x \in \left[0 , 3 \pi\right]$

$\therefore x = 0 , \frac{\pi}{2} , \pi , \frac{3 \pi}{2} , \ldots . .$
and
$= - \frac{\pi}{2} , - \pi , - \frac{3 \pi}{2} , \ldots .$

In general: $f ' \left(x\right) = 0$ for $x = \frac{n \pi}{2} \forall n \in \mathbb{Z}$

$f {\left(x\right)}_{\max} = 3 \cdot {\left(\pm 1\right)}^{2} = 3$

$f {\left(x\right)}_{\min} = 3 \cdot 0 = 0$

Hence critical points are:

$f {\left(x\right)}_{\max} = 3$ for $x = \frac{\left(2 n - 1\right) \pi}{2} \forall n \in \mathbb{Z}$

and

$f {\left(x\right)}_{\min} = 0$ for $x = \left(n \pi\right) \forall n \in \mathbb{Z}$

$\therefore$ Local max: $3$, Local min: $0$

These points can be seen on the graph of $f \left(x\right)$ below:

graph{3*(sinx)^2 [-10, 10, -5, 5]}