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

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

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

extreme points

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

Maximum (πn, 0), Minimum (\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:

Decreasing

: πn < x < \frac{π}{2} + πn,

Increasing

: \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

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

Maximum (πn, 0)

Plug the extreme point

x = \frac{π}{2} + πn

into

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

Minimum (\frac{π}{2} + πn, - 2)

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

f (x) = \frac{1}{4} x^{4} + \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 4

f (x) = x y^{2} = 9

f (x) = sin (θ)

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

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

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