What is the extreme f(x)=4cos(3x)+5,(0,2pi) ?

The extreme f(x)=4cos(3x)+5,(0,2pi) is

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extreme f(x)=4cos(3x)+5,(0,2pi)

extreme points

f (x) = 4 cos (3 x) + 5, (0, 2 π)

Minimum (\frac{π}{3}, 1), Maximum (\frac{2 π}{3}, 9), Minimum (π, 1), Maximum (\frac{4 π}{3}, 9), Minimum (\frac{5 π}{3}, 1)

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) + 5, (0, 2 π) : 0 < x < 2 π

Find intervals:

Decreasing

: 0 < x < \frac{π}{3},

Increasing

: \frac{π}{3} < x < \frac{2 π}{3},

Decreasing

: \frac{2 π}{3} < x < π,

Increasing

: π < x < \frac{4 π}{3},

Decreasing

: \frac{4 π}{3} < x < \frac{5 π}{3},

Increasing

: \frac{5 π}{3} < x < 2 π

Plug the extreme point

x = \frac{π}{3}

into

4 cos (3 x) + 5 \Rightarrow y = 1

Minimum (\frac{π}{3}, 1)

Plug the extreme point

x = \frac{2 π}{3}

into

4 cos (3 x) + 5 \Rightarrow y = 9

Maximum (\frac{2 π}{3}, 9)

Plug the extreme point

x = π

into

4 cos (3 x) + 5 \Rightarrow y = 1

Minimum (π, 1)

Plug the extreme point

x = \frac{4 π}{3}

into

4 cos (3 x) + 5 \Rightarrow y = 9

Maximum (\frac{4 π}{3}, 9)

Plug the extreme point

x = \frac{5 π}{3}

into

4 cos (3 x) + 5 \Rightarrow y = 1

Minimum (\frac{5 π}{3}, 1)

Minimum (\frac{π}{3}, 1), Maximum (\frac{2 π}{3}, 9), Minimum (π, 1), Maximum (\frac{4 π}{3}, 9), Minimum (\frac{5 π}{3}, 1)

f (x, y) = 14 x y - x^{3} - 7 y^{2}

f (x, y) = x^{2} + x y + y^{2} - 25 y + 208

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

y = \frac{x ( z + 2 )}{7}

x^{2} + 3

What is the extreme f(x)=4cos(3x)+5,(0,2pi) ?
The extreme f(x)=4cos(3x)+5,(0,2pi) is