What is the extreme f(x)=sin(x)-sin^2(x) ?

The extreme f(x)=sin(x)-sin^2(x) is

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extreme f(x)=sin(x)-sin^2(x)

extreme

Maximum (\frac{π}{6} + 2 πn, \frac{1}{4}), Minimum (\frac{π}{2} + 2 πn, 0), Maximum (\frac{5 π}{6} + 2 πn, \frac{1}{4}), Minimum (\frac{3 π}{2} + 2 πn, - 2)

Solution steps

Find the critical points:

x = \frac{π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Domain of

sin (x) - sin^{2} (x) : - \infty < x < \infty

Find intervals:

Increasing

: 2 πn \leq x < \frac{π}{6} + 2 πn,

Decreasing

: \frac{π}{6} + 2 πn < x < \frac{π}{2} + 2 πn,

Increasing

: \frac{π}{2} + 2 πn < x < \frac{5 π}{6} + 2 πn,

Decreasing

: \frac{5 π}{6} + 2 πn < x < \frac{3 π}{2} + 2 πn,

Increasing

: \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Plug the extreme point

x = \frac{π}{6} + 2 πn

into

sin (x) - sin^{2} (x) \Rightarrow y = \frac{1}{4}

Maximum (\frac{π}{6} + 2 πn, \frac{1}{4})

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

sin (x) - sin^{2} (x) \Rightarrow y = 0

Minimum (\frac{π}{2} + 2 πn, 0)

Plug the extreme point

x = \frac{5 π}{6} + 2 πn

into

sin (x) - sin^{2} (x) \Rightarrow y = \frac{1}{4}

Maximum (\frac{5 π}{6} + 2 πn, \frac{1}{4})

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

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

Minimum (\frac{3 π}{2} + 2 πn, - 2)

Maximum (\frac{π}{6} + 2 πn, \frac{1}{4}), Minimum (\frac{π}{2} + 2 πn, 0), Maximum (\frac{5 π}{6} + 2 πn, \frac{1}{4}), Minimum (\frac{3 π}{2} + 2 πn, - 2)