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

The extreme f(x)=2x-4sin(x),0<= x<= 2pi is

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extreme f(x)=2x-4sin(x),0<= x<= 2pi

extreme points

f (x) = 2 x - 4 sin (x), 0 \leq x \leq 2 π

Maximum (0, 0), Minimum (\frac{π}{3}, \frac{2 π}{3} - 23), Maximum (\frac{5 π}{3}, \frac{10 π}{3} + 23), Minimum (2 π, 4 π)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{3}, x = \frac{5 π}{3}, x = 2 π

Domain of

2 x - 4 sin (x), 0 \leq x \leq 2 π : 0 \leq x \leq 2 π

Find intervals:

Decreasing

: 0 < x < \frac{π}{3},

Increasing

: \frac{π}{3} < x < \frac{5 π}{3},

Decreasing

: \frac{5 π}{3} < x < 2 π

Plug the extreme point

x = 0

into

2 x - 4 sin (x) \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = \frac{π}{3}

into

2 x - 4 sin (x) \Rightarrow y = \frac{2 π}{3} - 23

Minimum (\frac{π}{3}, \frac{2 π}{3} - 23)

Plug the extreme point

x = \frac{5 π}{3}

into

2 x - 4 sin (x) \Rightarrow y = \frac{10 π}{3} + 23

Maximum (\frac{5 π}{3}, \frac{10 π}{3} + 23)

Plug the extreme point

x = 2 π

into

2 x - 4 sin (x) \Rightarrow y = 4 π

Minimum (2 π, 4 π)

Maximum (0, 0), Minimum (\frac{π}{3}, \frac{2 π}{3} - 23), Maximum (\frac{5 π}{3}, \frac{10 π}{3} + 23), Minimum (2 π, 4 π)

f (x) = \frac{x ^{3}}{3} - 2 x^{2} + 4, - 2 \leq x \leq 1

f (x, y) = x^{4} + 6 y^{2} - 4 x y^{3} - 1

3 (x - 2)

f (x) = 2 sin (x) + 2 cos (x)

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

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