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

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

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

extreme points

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

Minimum (0, 0), Maximum (\frac{π}{3}, 3 - \frac{π}{3}), Minimum (\frac{5 π}{3}, - 3 - \frac{5 π}{3}), Maximum (2 π, - 2 π)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{3}, x = \frac{5 π}{3}, x = 2 π

Domain of

2 sin (x) - x, 0 \leq x \leq 2 π : 0 \leq x \leq 2 π

Find intervals:

Increasing

: 0 < x < \frac{π}{3},

Decreasing

: \frac{π}{3} < x < \frac{5 π}{3},

Increasing

: \frac{5 π}{3} < x < 2 π

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = \frac{π}{3}

into

2 sin (x) - x \Rightarrow y = 3 - \frac{π}{3}

Maximum (\frac{π}{3}, 3 - \frac{π}{3})

Plug the extreme point

x = \frac{5 π}{3}

into

2 sin (x) - x \Rightarrow y = - 3 - \frac{5 π}{3}

Minimum (\frac{5 π}{3}, - 3 - \frac{5 π}{3})

Plug the extreme point

x = 2 π

into

2 sin (x) - x \Rightarrow y = - 2 π

Maximum (2 π, - 2 π)

Minimum (0, 0), Maximum (\frac{π}{3}, 3 - \frac{π}{3}), Minimum (\frac{5 π}{3}, - 3 - \frac{5 π}{3}), Maximum (2 π, - 2 π)

f (x) = 8 + 6 x - 7 x^{2} - 4 x^{3}

f (x) = \frac{20}{3 ( 2 + x ) ^{3}}

f (x) = \frac{- x - 5}{x ^{2} + 16 x + 64}

y = 2 sin (∣ x ∣), - 2 π \leq x \leq 2 π

f (x) = x^{3} \cdot ln (x)

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