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

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

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extreme f(x)=cos(2x)

extreme points

f (x) = cos (2 x)

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

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

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

cos (2 x) \Rightarrow y = 1

Maximum (πn, 1)

Plug the extreme point

x = \frac{π}{2} + πn

into

cos (2 x) \Rightarrow y = - 1

Minimum (\frac{π}{2} + πn, - 1)

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

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

g (x) = 3 + x - 4

y = x^{2} - 9

9 x + 12 y = 12

y = x + 6

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