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

The extreme f(x)=x+2cos(x),0<= x<= 2pi is

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extreme f(x)=x+2cos(x),0<= x<= 2pi

extreme points

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

Minimum (0, 2), Maximum (\frac{π}{6}, \frac{π}{6} + 3), Minimum (\frac{5 π}{6}, \frac{5 π}{6} - 3), Maximum (2 π, 2 π + 2)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{6}, x = \frac{5 π}{6}, x = 2 π

Domain of

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

Find intervals:

Increasing

: 0 < x < \frac{π}{6},

Decreasing

: \frac{π}{6} < x < \frac{5 π}{6},

Increasing

: \frac{5 π}{6} < x < 2 π

Plug the extreme point

x = 0

into

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

Minimum (0, 2)

Plug the extreme point

x = \frac{π}{6}

into

x + 2 cos (x) \Rightarrow y = \frac{π}{6} + 3

Maximum (\frac{π}{6}, \frac{π}{6} + 3)

Plug the extreme point

x = \frac{5 π}{6}

into

x + 2 cos (x) \Rightarrow y = \frac{5 π}{6} - 3

Minimum (\frac{5 π}{6}, \frac{5 π}{6} - 3)

Plug the extreme point

x = 2 π

into

x + 2 cos (x) \Rightarrow y = 2 π + 2

Maximum (2 π, 2 π + 2)

Minimum (0, 2), Maximum (\frac{π}{6}, \frac{π}{6} + 3), Minimum (\frac{5 π}{6}, \frac{5 π}{6} - 3), Maximum (2 π, 2 π + 2)

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

f (x) = 8 sin (2 x), - 2 π \leq x \leq 2 π

f (x) = \frac{( 4 x ^{3} )}{3} - 4 x^{2} - 60 x + 35

f (x, y) = 12 x y + 4321

f (x) = 5 x + 9

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