What is the extreme f(x)=3cos(2x)+4 ?

The extreme f(x)=3cos(2x)+4 is Maximum(pin,7),Minimum(pi/2+pin,1)

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extreme f(x)=3cos(2x)+4

extreme points

f (x) = 3 cos (2 x) + 4

Maximum (πn, 7), Minimum (\frac{π}{2} + πn, 1)

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

3 cos (2 x) + 4 : - \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

3 cos (2 x) + 4 \Rightarrow y = 7

Maximum (πn, 7)

Plug the extreme point

x = \frac{π}{2} + πn

into

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

Minimum (\frac{π}{2} + πn, 1)

Maximum (πn, 7), Minimum (\frac{π}{2} + πn, 1)

f (x) = 4 x^{3} - 12 x + 7

f (x) = 3 x + 6 x^{- 1}

f (x, y) = - x^{3} + y^{3} + 15 x y + 1

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

f (x) = - 3 x^{2} + 10 x - 10

What is the extreme f(x)=3cos(2x)+4 ?
The extreme f(x)=3cos(2x)+4 is Maximum(pin,7),Minimum(pi/2+pin,1)