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),Saddle(pi,pi),Maximum(2pi,2pi)

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

extreme

Minimum (0, 0), Saddle (π, π), Maximum (2 π, 2 π)

Solution steps

Find the critical points:

x = 0, x = π, 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 < π,

Increasing

: π < x < 2 π

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = π

into

x - sin (- x) \Rightarrow y = π

Saddle (π, π)

Plug the extreme point

x = 2 π

into

x - sin (- x) \Rightarrow y = 2 π

Maximum (2 π, 2 π)

Minimum (0, 0), Saddle (π, π), Maximum (2 π, 2 π)

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),Saddle(pi,pi),Maximum(2pi,2pi)