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

The extreme f(x)=sin^2(x) is Minimum(pin,0),Maximum(pi/2+pin,1)

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

extreme points

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

Minimum (πn, 0), Maximum (\frac{π}{2} + πn, 1)

Solution steps

Find the critical points:

x = πn, x = \frac{π}{2} + πn

Domain of

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

Find intervals:

Increasing

: πn < x < \frac{π}{2} + πn,

Decreasing

: \frac{π}{2} + πn < x < π + πn

Plug the extreme point

x = πn

into

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

Minimum (πn, 0)

Plug the extreme point

x = \frac{π}{2} + πn

into

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

Maximum (\frac{π}{2} + πn, 1)

Minimum (πn, 0), Maximum (\frac{π}{2} + πn, 1)

y = x - 4 + 3

\frac{x ^{2} - 2}{x - 2}

f (x) = - \frac{2}{5} x + 6

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

f (x) = 5 x^{2} - 4 x^{6} + x^{4}

What is the extreme f(x)=sin^2(x) ?
The extreme f(x)=sin^2(x) is Minimum(pin,0),Maximum(pi/2+pin,1)