What is the extreme f(x)=-4cos(3x)-4,(0,2pi) ?

The extreme f(x)=-4cos(3x)-4,(0,2pi) is

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extreme f(x)=-4cos(3x)-4,(0,2pi)

extreme points

f (x) = - 4 cos (3 x) - 4, (0, 2 π)

Maximum (\frac{π}{3}, 0), Minimum (\frac{2 π}{3}, - 8), Maximum (π, 0), Minimum (\frac{4 π}{3}, - 8), Maximum (\frac{5 π}{3}, 0)

Solution steps

Find the critical points:

x = \frac{π}{3}, x = \frac{2 π}{3}, x = π, x = \frac{4 π}{3}, x = \frac{5 π}{3}

Domain of

- 4 cos (3 x) - 4, (0, 2 π) : 0 < x < 2 π

Find intervals:

Increasing

: 0 < x < \frac{π}{3},

Decreasing

: \frac{π}{3} < x < \frac{2 π}{3},

Increasing

: \frac{2 π}{3} < x < π,

Decreasing

: π < x < \frac{4 π}{3},

Increasing

: \frac{4 π}{3} < x < \frac{5 π}{3},

Decreasing

: \frac{5 π}{3} < x < 2 π

Plug the extreme point

x = \frac{π}{3}

into

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

Maximum (\frac{π}{3}, 0)

Plug the extreme point

x = \frac{2 π}{3}

into

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

Minimum (\frac{2 π}{3}, - 8)

Plug the extreme point

x = π

into

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

Maximum (π, 0)

Plug the extreme point

x = \frac{4 π}{3}

into

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

Minimum (\frac{4 π}{3}, - 8)

Plug the extreme point

x = \frac{5 π}{3}

into

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

Maximum (\frac{5 π}{3}, 0)

Maximum (\frac{π}{3}, 0), Minimum (\frac{2 π}{3}, - 8), Maximum (π, 0), Minimum (\frac{4 π}{3}, - 8), Maximum (\frac{5 π}{3}, 0)

f (t) = 5 - 6 u (t - 2) + 3 u (t - 5)

h (t, v) = - 16 t^{2} + v t + k

f (x, y) = 4 x - 8 x y + 2 y + 1

+ 5^{x} - 4

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

What is the extreme f(x)=-4cos(3x)-4,(0,2pi) ?
The extreme f(x)=-4cos(3x)-4,(0,2pi) is