What is the extreme f(x)=4cos(2x)+1 ?

The extreme f(x)=4cos(2x)+1 is Maximum(pin,5),Minimum(pi/2+pin,-3)

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extreme f(x)=4cos(2x)+1

extreme

Maximum (πn, 5), Minimum (\frac{π}{2} + πn, - 3)

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

4 cos (2 x) + 1 : - \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

4 cos (2 x) + 1 \Rightarrow y = 5

Maximum (πn, 5)

Plug the extreme point

x = \frac{π}{2} + πn

into

4 cos (2 x) + 1 \Rightarrow y = - 3

Minimum (\frac{π}{2} + πn, - 3)

Maximum (πn, 5), Minimum (\frac{π}{2} + πn, - 3)

What is the extreme f(x)=4cos(2x)+1 ?
The extreme f(x)=4cos(2x)+1 is Maximum(pin,5),Minimum(pi/2+pin,-3)