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

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

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

extreme points

f (x) = \frac{x}{2} - sin (x), - π \leq x \leq π

Minimum (- π, - \frac{π}{2}), Maximum (- \frac{π}{3}, - \frac{π}{6} + \frac{3}{2}), Minimum (\frac{π}{3}, \frac{π}{6} - \frac{3}{2}), Maximum (π, \frac{π}{2})

Solution steps

Find the critical points:

x = - π, x = - \frac{π}{3}, x = \frac{π}{3}, x = π

Domain of

\frac{x}{2} - sin (x), - π \leq x \leq π : - π \leq x \leq π

Find intervals:

Increasing

: - π < x < - \frac{π}{3},

Decreasing

: - \frac{π}{3} < x < \frac{π}{3},

Increasing

: \frac{π}{3} < x < π

Plug the extreme point

x = - π

into

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

Minimum (- π, - \frac{π}{2})

Plug the extreme point

x = - \frac{π}{3}

into

\frac{x}{2} - sin (x) \Rightarrow y = - \frac{π}{6} + \frac{3}{2}

Maximum (- \frac{π}{3}, - \frac{π}{6} + \frac{3}{2})

Plug the extreme point

x = \frac{π}{3}

into

\frac{x}{2} - sin (x) \Rightarrow y = \frac{π}{6} - \frac{3}{2}

Minimum (\frac{π}{3}, \frac{π}{6} - \frac{3}{2})

Plug the extreme point

x = π

into

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

Maximum (π, \frac{π}{2})

Minimum (- π, - \frac{π}{2}), Maximum (- \frac{π}{3}, - \frac{π}{6} + \frac{3}{2}), Minimum (\frac{π}{3}, \frac{π}{6} - \frac{3}{2}), Maximum (π, \frac{π}{2})

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

f (x) = x (20 - 2 x) (12 - 2 x)

f (x) = 5 x^{2} - 40 ln (x)

y = x^{3} - 6 x^{2} + 2

f (x) = x^{3} - 3 x + 9

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