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

The extreme f(x)=4x+4sin(x),0<= x<= 2pi is Minimum(0,0),Saddle(pi,4pi),Maximum(2pi,8pi)

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

extreme points

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

Minimum (0, 0), Saddle (π, 4 π), Maximum (2 π, 8 π)

Solution steps

Find the critical points:

x = 0, x = π, x = 2 π

Domain of

4 x + 4 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

4 x + 4 sin (x) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = π

into

4 x + 4 sin (x) \Rightarrow y = 4 π

Saddle (π, 4 π)

Plug the extreme point

x = 2 π

into

4 x + 4 sin (x) \Rightarrow y = 8 π

Maximum (2 π, 8 π)

Minimum (0, 0), Saddle (π, 4 π), Maximum (2 π, 8 π)

f (x, y) = x^{2} + 7 x y + y^{2}

f (x) = 2 x^{2} - 16 ln (x)

f (x, y) = x 1 + y^{3}

f (x) = \frac{1 + x}{1 + x ^{2}}

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

What is the extreme f(x)=4x+4sin(x),0<= x<= 2pi ?
The extreme f(x)=4x+4sin(x),0<= x<= 2pi is Minimum(0,0),Saddle(pi,4pi),Maximum(2pi,8pi)