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

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

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

extreme points

f (x) = cos (3 x) - 2, (0, 2 π)

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

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

cos (3 x) - 2, (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

cos (3 x) - 2 \Rightarrow y = - 3

Minimum (\frac{π}{3}, - 3)

Plug the extreme point

x = \frac{2 π}{3}

into

cos (3 x) - 2 \Rightarrow y = - 1

Maximum (\frac{2 π}{3}, - 1)

Plug the extreme point

x = π

into

cos (3 x) - 2 \Rightarrow y = - 3

Minimum (π, - 3)

Plug the extreme point

x = \frac{4 π}{3}

into

cos (3 x) - 2 \Rightarrow y = - 1

Maximum (\frac{4 π}{3}, - 1)

Plug the extreme point

x = \frac{5 π}{3}

into

cos (3 x) - 2 \Rightarrow y = - 3

Minimum (\frac{5 π}{3}, - 3)

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

f (x) = \frac{1}{x + 1}

f (x) = - 3 sin (2 x) - 4

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

f (x) = 2 sin^{2} (x), 0 \leq x \leq π

f (x) = 4 cos (3 x) + 5, (0, 2 π)

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