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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = - \frac{π}{2}, x = - \frac{π}{6}, x = \frac{π}{6}, x = \frac{π}{2}

Domain of

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

Find intervals:

Decreasing

: - \frac{π}{2} < x < - \frac{π}{6},

Increasing

: - \frac{π}{6} < x < \frac{π}{6},

Decreasing

: \frac{π}{6} < x < \frac{π}{2}

Plug the extreme point

x = - \frac{π}{2}

into

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

Maximum (- \frac{π}{2}, \frac{π}{2})

Plug the extreme point

x = - \frac{π}{6}

into

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

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

Plug the extreme point

x = \frac{π}{6}

into

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

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

Plug the extreme point

x = \frac{π}{2}

into

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

Minimum (\frac{π}{2}, - \frac{π}{2})

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

f (x) = 5 x^{4} - x^{5} + 8

f (x) = 5 x^{4} - x^{5} + 7

f (x) = 5 x^{4} - x^{5} + 4

17 r^{2} - r^{3}

f (x, y) = 7 - x^{4} + 2 x^{2} - y^{2}

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