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

The extreme f(x)=3cos(2x)+3x is

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

extreme points

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

Maximum (\frac{π}{12} + πn, 3 cos (2 (\frac{π}{12} + πn)) + 3 (\frac{π}{12} + πn)), Minimum (\frac{5 π}{12} + πn, 3 cos (2 (\frac{5 π}{12} + πn)) + 3 (\frac{5 π}{12} + πn))

Solution steps

Find the critical points:

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Domain of

3 cos (2 x) + 3 x : - \infty < x < \infty

Find intervals:

Increasing

: πn \leq x < \frac{π}{12} + πn,

Decreasing

: \frac{π}{12} + πn < x < \frac{5 π}{12} + πn,

Increasing

: \frac{5 π}{12} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{12} + πn

into

3 cos (2 x) + 3 x \Rightarrow y = 3 cos (2 (\frac{π}{12} + πn)) + 3 (\frac{π}{12} + πn)

Maximum (\frac{π}{12} + πn, 3 cos (2 (\frac{π}{12} + πn)) + 3 (\frac{π}{12} + πn))

Plug the extreme point

x = \frac{5 π}{12} + πn

into

3 cos (2 x) + 3 x \Rightarrow y = 3 cos (2 (\frac{5 π}{12} + πn)) + 3 (\frac{5 π}{12} + πn)

Minimum (\frac{5 π}{12} + πn, 3 cos (2 (\frac{5 π}{12} + πn)) + 3 (\frac{5 π}{12} + πn))

Maximum (\frac{π}{12} + πn, 3 cos (2 (\frac{π}{12} + πn)) + 3 (\frac{π}{12} + πn)), Minimum (\frac{5 π}{12} + πn, 3 cos (2 (\frac{5 π}{12} + πn)) + 3 (\frac{5 π}{12} + πn))

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