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

The extreme f(x)=9cos(x),0<= x<= 2pi is Maximum(0,9),Minimum(pi,-9),Maximum(2pi,9)

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

extreme

Maximum (0, 9), Minimum (π, - 9), Maximum (2 π, 9)

Solution steps

Find the critical points:

x = 0, x = π, x = 2 π

Domain of

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

Find intervals:

Decreasing

: 0 < x < π,

Increasing

: π < x < 2 π

Plug the extreme point

x = 0

into

9 cos (x) \Rightarrow y = 9

Maximum (0, 9)

Plug the extreme point

x = π

into

9 cos (x) \Rightarrow y = - 9

Minimum (π, - 9)

Plug the extreme point

x = 2 π

into

9 cos (x) \Rightarrow y = 9

Maximum (2 π, 9)

Maximum (0, 9), Minimum (π, - 9), Maximum (2 π, 9)

What is the extreme f(x)=9cos(x),0<= x<= 2pi ?
The extreme f(x)=9cos(x),0<= x<= 2pi is Maximum(0,9),Minimum(pi,-9),Maximum(2pi,9)