What is the extreme f(x)=9x+9sin(x),0<= x<= 2pi ?

The extreme f(x)=9x+9sin(x),0<= x<= 2pi is Minimum(0,0),Saddle(pi,9pi),Maximum(2pi,18pi)

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extreme f(x)=9x+9sin(x),0<= x<= 2pi

extreme points

f (x) = 9 x + 9 sin (x), 0 \leq x \leq 2 π

Minimum (0, 0), Saddle (π, 9 π), Maximum (2 π, 18 π)

Solution steps

Find the critical points:

x = 0, x = π, x = 2 π

Domain of

9 x + 9 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

9 x + 9 sin (x) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = π

into

9 x + 9 sin (x) \Rightarrow y = 9 π

Saddle (π, 9 π)

Plug the extreme point

x = 2 π

into

9 x + 9 sin (x) \Rightarrow y = 18 π

Maximum (2 π, 18 π)

Minimum (0, 0), Saddle (π, 9 π), Maximum (2 π, 18 π)

f (x) = x e^{- 4 x^{2}}

y = x^{\frac{2}{5}} (x + 3)

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

x + 2 cos (x)

f (x) = - 2 x + 3 ln (4 x)

What is the extreme f(x)=9x+9sin(x),0<= x<= 2pi ?
The extreme f(x)=9x+9sin(x),0<= x<= 2pi is Minimum(0,0),Saddle(pi,9pi),Maximum(2pi,18pi)