What is the extreme f(x)=sin^2(x),0<= x<= pi ?

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

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extreme f(x)=sin^2(x),0<= x<= pi

extreme points

f (x) = sin^{2} (x), 0 \leq x \leq π

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

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = π

Domain of

sin^{2} (x), 0 \leq x \leq π : 0 \leq x \leq π

Find intervals:

Increasing

: 0 < x < \frac{π}{2},

Decreasing

: \frac{π}{2} < x < π

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = \frac{π}{2}

into

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

Maximum (\frac{π}{2}, 1)

Plug the extreme point

x = π

into

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

Minimum (π, 0)

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

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

(x + 3)^{2} - 1

f (x) = \frac{x ^{2} + 5 x}{25 - x ^{2}}

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

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

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