What is the extreme y=sqrt(3)x+2cos(x),0<= x<= 2pi ?

The extreme y=sqrt(3)x+2cos(x),0<= x<= 2pi is

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extreme y=sqrt(3)x+2cos(x),0<= x<= 2pi

extreme points

y = 3 x + 2 cos (x), 0 \leq x \leq 2 π

Minimum (0, 2), Maximum (\frac{π}{3}, \frac{π}{3} + 1), Minimum (\frac{2 π}{3}, \frac{2 π}{3} - 1), Maximum (2 π, 23 π + 2)

Solution steps

Find the critical points:

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

Domain of

3 x + 2 cos (x), 0 \leq x \leq 2 π : 0 \leq x \leq 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 < 2 π

Plug the extreme point

x = 0

into

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

Minimum (0, 2)

Plug the extreme point

x = \frac{π}{3}

into

3 x + 2 cos (x) \Rightarrow y = \frac{π}{3} + 1

Maximum (\frac{π}{3}, \frac{π}{3} + 1)

Plug the extreme point

x = \frac{2 π}{3}

into

3 x + 2 cos (x) \Rightarrow y = \frac{2 π}{3} - 1

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

Plug the extreme point

x = 2 π

into

3 x + 2 cos (x) \Rightarrow y = 23 π + 2

Maximum (2 π, 23 π + 2)

Minimum (0, 2), Maximum (\frac{π}{3}, \frac{π}{3} + 1), Minimum (\frac{2 π}{3}, \frac{2 π}{3} - 1), Maximum (2 π, 23 π + 2)

f (x) = 2 x - \frac{108}{x}

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

f (x) = \frac{x}{x - 4}

f (x) = x^{2} + x y + y^{2} - 19 y + 120

f (x) = x^{2} + x - 12

What is the extreme y=sqrt(3)x+2cos(x),0<= x<= 2pi ?
The extreme y=sqrt(3)x+2cos(x),0<= x<= 2pi is