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

The extreme f(x)=-4cos(3x)-4 is Minimum((2pi)/3 n,-8),Maximum(pi/3+(2pi)/3 n,0)

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extreme f(x)=-4cos(3x)-4

extreme points

f (x) = - 4 cos (3 x) - 4

Minimum (\frac{2 π}{3} n, - 8), Maximum (\frac{π}{3} + \frac{2 π}{3} n, 0)

Solution steps

Find the critical points:

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

Domain of

- 4 cos (3 x) - 4 : - \infty < x < \infty

Find intervals:

Increasing

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

Decreasing

: \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

- 4 cos (3 x) - 4 \Rightarrow y = - 8

Minimum (\frac{2 π}{3} n, - 8)

Plug the extreme point

x = \frac{π}{3} + \frac{2 π}{3} n

into

- 4 cos (3 x) - 4 \Rightarrow y = 0

Maximum (\frac{π}{3} + \frac{2 π}{3} n, 0)

Minimum (\frac{2 π}{3} n, - 8), Maximum (\frac{π}{3} + \frac{2 π}{3} n, 0)

f (x, y) = 200 y^{2} + x^{2} - x^{2} y

2 x^{4} - 16 x^{2} + 3

F (Y, Z) = Y + Y Z

f (x, y) = 14 x^{2} - 2 x^{3} + 4 x y

f (x) = \frac{2}{3} x^{3} + 2 x^{2} + 6 x - 5

What is the extreme f(x)=-4cos(3x)-4 ?
The extreme f(x)=-4cos(3x)-4 is Minimum((2pi)/3 n,-8),Maximum(pi/3+(2pi)/3 n,0)