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

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

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extreme f(x)=cos(3x)

extreme points

f (x) = cos (3 x)

Maximum (\frac{2 π}{3} n, 1), Minimum (\frac{π}{3} + \frac{2 π}{3} n, - 1)

Solution steps

Find the critical points:

x = \frac{2 π}{3} n, x = \frac{π}{3} + \frac{2 π}{3} n

Domain of

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

Find intervals:

Decreasing

: \frac{2 π}{3} n < x < \frac{π}{3} + \frac{2 π}{3} n,

Increasing

: \frac{π}{3} + \frac{2 π}{3} n < x < \frac{2 π}{3} + \frac{2 π}{3} n

Plug the extreme point

x = \frac{2 π}{3} n

into

cos (3 x) \Rightarrow y = 1

Maximum (\frac{2 π}{3} n, 1)

Plug the extreme point

x = \frac{π}{3} + \frac{2 π}{3} n

into

cos (3 x) \Rightarrow y = - 1

Minimum (\frac{π}{3} + \frac{2 π}{3} n, - 1)

Maximum (\frac{2 π}{3} n, 1), Minimum (\frac{π}{3} + \frac{2 π}{3} n, - 1)

f (x) = 2 x^{3} - 3 x^{2} + 6

f (x, y) = x^{2} + 9 y^{2}

f (x) = 2 x + \frac{8}{x ^{2}} + 5, x > 0

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

f (x, y) = \frac{1}{1 - x ^{2} - y ^{2}}

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