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

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

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

extreme points

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

Minimum (0, 0), Maximum (2 π, 2 π)

Solution steps

Find the critical points:

x = 0, x = 2 π

Domain of

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

Find intervals:

Increasing

: 0 < x < 2 π

Plug the extreme point

x = 0

into

x - sin (x) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 2 π

into

x - sin (x) \Rightarrow y = 2 π

Maximum (2 π, 2 π)

Minimum (0, 0), Maximum (2 π, 2 π)

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

y = x^{2} - 2 x - 8

f (x, y) = 5^{2} x^{4} + 5^{2} y^{4} - 4 x y + 14

f (x) = x^{3} - 3 x^{2} + x - 1

f (x) = x^{3} + 3 x^{2} + 5

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