What is the extreme f(x)=9sin^2(x)+9sin(x) ?

The extreme f(x)=9sin^2(x)+9sin(x) is

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extreme f(x)=9sin^2(x)+9sin(x)

extreme points

f (x) = 9 sin^{2} (x) + 9 sin (x)

Maximum (\frac{π}{2} + 2 πn, 18), Minimum (\frac{7 π}{6} + 2 πn, - \frac{9}{4}), Maximum (\frac{3 π}{2} + 2 πn, 0), Minimum (\frac{11 π}{6} + 2 πn, - \frac{9}{4})

Solution steps

Find the critical points:

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{11 π}{6} + 2 πn

Domain of

9 sin^{2} (x) + 9 sin (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Decreasing

: \frac{π}{2} + 2 πn < x < \frac{7 π}{6} + 2 πn,

Increasing

: \frac{7 π}{6} + 2 πn < x < \frac{3 π}{2} + 2 πn,

Decreasing

: \frac{3 π}{2} + 2 πn < x < \frac{11 π}{6} + 2 πn,

Increasing

: \frac{11 π}{6} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

9 sin^{2} (x) + 9 sin (x) \Rightarrow y = 18

Maximum (\frac{π}{2} + 2 πn, 18)

Plug the extreme point

x = \frac{7 π}{6} + 2 πn

into

9 sin^{2} (x) + 9 sin (x) \Rightarrow y = - \frac{9}{4}

Minimum (\frac{7 π}{6} + 2 πn, - \frac{9}{4})

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

9 sin^{2} (x) + 9 sin (x) \Rightarrow y = 0

Maximum (\frac{3 π}{2} + 2 πn, 0)

Plug the extreme point

x = \frac{11 π}{6} + 2 πn

into

9 sin^{2} (x) + 9 sin (x) \Rightarrow y = - \frac{9}{4}

Minimum (\frac{11 π}{6} + 2 πn, - \frac{9}{4})

Maximum (\frac{π}{2} + 2 πn, 18), Minimum (\frac{7 π}{6} + 2 πn, - \frac{9}{4}), Maximum (\frac{3 π}{2} + 2 πn, 0), Minimum (\frac{11 π}{6} + 2 πn, - \frac{9}{4})

h (x) = - 6 x^{3} + 18 x^{2} + 3

F (x) = \frac{2 x}{x ^{2} + 16}, - 8 \leq x \leq 8

- (\frac{9}{2})^{2} - 7^{2} + \frac{9}{2} - 14 + 3

x (8 - x^{2})

f (x) = f (x) = - 8 x^{2} - 144 x + 6

What is the extreme f(x)=9sin^2(x)+9sin(x) ?
The extreme f(x)=9sin^2(x)+9sin(x) is