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

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

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extreme 2sin^2(x)

extreme points

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, y) = x^{2} + x - 3 x y + y^{3} - 5

y = 5 - 2 x^{2}, - 4 \leq x \leq 2

f (x) = x^{3} - \frac{7}{2} x^{2} + 2 x - 5

f (x) = (t^{2} - 4)^{3}

\frac{x}{x ^{2} + 16}, 0 \leq x \leq 8

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