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

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

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

extreme points

f (x) = sin^{2} (x) - sin (x)

Minimum (\frac{π}{6} + 2 πn, - \frac{1}{4}), Maximum (\frac{π}{2} + 2 πn, 0), Minimum (\frac{5 π}{6} + 2 πn, - \frac{1}{4}), Maximum (\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^{2} (x) - sin (x) : - \infty < x < \infty

Find intervals:

Decreasing

: 2 πn \leq x < \frac{π}{6} + 2 πn,

Increasing

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

Decreasing

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

Increasing

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

Decreasing

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

Plug the extreme point

x = \frac{π}{6} + 2 πn

into

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

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

Plug the extreme point

x = \frac{π}{2} + 2 πn

into

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

Maximum (\frac{π}{2} + 2 πn, 0)

Plug the extreme point

x = \frac{5 π}{6} + 2 πn

into

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

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

Plug the extreme point

x = \frac{3 π}{2} + 2 πn

into

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

Maximum (\frac{3 π}{2} + 2 πn, 2)

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

f (x) = 4 x^{3} + 9 x^{2} - 12 x + 7

h (v, t) = v t - 16 t^{2}

f (x) = - x^{2} - 2 x + 3

f (x) = x (x + a)

x^{2} - 2 x y + 2 y