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

The extreme f(x)=2cos(2x)-2 is Maximum(pin,0),Minimum(pi/2+pin,-4)

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

extreme points

f (x) = 2 cos (2 x) - 2

Maximum (πn, 0), Minimum (\frac{π}{2} + πn, - 4)

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

2 cos (2 x) - 2 : - \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) - 2 \Rightarrow y = 0

Maximum (πn, 0)

Plug the extreme point

x = \frac{π}{2} + πn

into

2 cos (2 x) - 2 \Rightarrow y = - 4

Minimum (\frac{π}{2} + πn, - 4)

Maximum (πn, 0), Minimum (\frac{π}{2} + πn, - 4)

2 x e^{- x^{2}}

R (x, y) = - 5 x^{2} - 8 y^{2} - 2 x y + 42 x + 102 y

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

4 x^{2} - 4 x

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

What is the extreme f(x)=2cos(2x)-2 ?
The extreme f(x)=2cos(2x)-2 is Maximum(pin,0),Minimum(pi/2+pin,-4)