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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

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

Maximum (πn, 2)

Plug the extreme point

x = \frac{π}{2} + πn

into

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

Minimum (\frac{π}{2} + πn, 0)

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

6 x^{\frac{2}{3}} - 4 x

f (x) = x^{2} e^{- 4 x}

f (x) = 4 x^{2} + 2 y^{2} - 2 x y - 10 y - 2 x

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

f (x) = 2 x^{2} - 6 x^{4}

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