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

The extreme f(x)=2sin^2(x) is Minimum(pin,0),Maximum(pi/2+pin,2)

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

extreme points

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

Minimum (πn, 0), Maximum (\frac{π}{2} + πn, 2)

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

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

Find intervals:

Increasing

: πn < x < \frac{π}{2} + πn,

Decreasing

: \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

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

Minimum (πn, 0)

Plug the extreme point

x = \frac{π}{2} + πn

into

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

Maximum (\frac{π}{2} + πn, 2)

Minimum (πn, 0), Maximum (\frac{π}{2} + πn, 2)

f (x) = 5 + 6 x - 8 x^{3}

f (x) = 2 x^{3} + 24 x + 5

f (x, y) = 3 - x^{4} + 2 x^{2} - y^{2}

f (x, y) = x y + y x

f (x) = x^{2} - 8 ln (x)

What is the extreme f(x)=2sin^2(x) ?
The extreme f(x)=2sin^2(x) is Minimum(pin,0),Maximum(pi/2+pin,2)