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

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

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

extreme points

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

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

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) - 2 : - \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) - 2 \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) - 2 \Rightarrow y = - 3

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

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

f (x) = - sin (x) - 4

f (x) = 2 x^{3} - x^{2} - 4 x + 8 (- 1)

f (x, y) = 2 x^{5} + x^{3} y^{2} + 6 x y^{4}

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

f (x, y) = x^{2} y^{3} - (x - 2 y)^{2} + 6 x^{4} y = 3 x y

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