What is the extreme f(x)=7x-7sin(x),0<= x<= 2pi ?

The extreme f(x)=7x-7sin(x),0<= x<= 2pi is Minimum(0,0),Maximum(2pi,14pi)

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extreme f(x)=7x-7sin(x),0<= x<= 2pi

extreme points

f (x) = 7 x - 7 sin (x), 0 \leq x \leq 2 π

Minimum (0, 0), Maximum (2 π, 14 π)

Solution steps

Find the critical points:

x = 0, x = 2 π

Domain of

7 x - 7 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

7 x - 7 sin (x) \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 2 π

into

7 x - 7 sin (x) \Rightarrow y = 14 π

Maximum (2 π, 14 π)

Minimum (0, 0), Maximum (2 π, 14 π)

f (x) = x^{4} + 12 x^{3} + 16 x^{2} + 4

f (x) = 4 x - x^{2} [1.5]

f (x) = 9 x y e^{- x^{2} - y^{2}}

f (x) = x^{3} - 12 x - 1

f (x, y) = 9 - x^{2} - 9 y^{2}

What is the extreme f(x)=7x-7sin(x),0<= x<= 2pi ?
The extreme f(x)=7x-7sin(x),0<= x<= 2pi is Minimum(0,0),Maximum(2pi,14pi)