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

The extreme f(x)=2sin(x)+sin^2(x) is Maximum(pi/2+2pin,3),Minimum((3pi)/2+2pin,-1)

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

extreme points

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

Maximum (\frac{π}{2} + 2 πn, 3), Minimum (\frac{3 π}{2} + 2 πn, - 1)

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{2} + 2 πn,

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

2 sin (x) + sin^{2} (x) \Rightarrow y = 3

Maximum (\frac{π}{2} + 2 πn, 3)

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

2 sin (x) + sin^{2} (x) \Rightarrow y = - 1

Minimum (\frac{3 π}{2} + 2 πn, - 1)

Maximum (\frac{π}{2} + 2 πn, 3), Minimum (\frac{3 π}{2} + 2 πn, - 1)

f (x) = x (1 - x)^{4}

x^{4} - 2 x^{2} + 9

\frac{x ^{2}}{x - 3}

x^{4} - 2 x^{2} + 2

f (x) = x^{\frac{3}{2}}, (- 1, 2)

What is the extreme f(x)=2sin(x)+sin^2(x) ?
The extreme f(x)=2sin(x)+sin^2(x) is Maximum(pi/2+2pin,3),Minimum((3pi)/2+2pin,-1)