What is the extreme f(x)=3sin^2(x),0<= x<= pi ?

The extreme f(x)=3sin^2(x),0<= x<= pi is Minimum(0,0),Maximum(pi/2 ,3),Minimum(pi,0)

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extreme f(x)=3sin^2(x),0<= x<= pi

extreme points

f (x) = 3 sin^{2} (x), 0 \leq x \leq π

Minimum (0, 0), Maximum (\frac{π}{2}, 3), Minimum (π, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = π

Domain of

3 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

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

Minimum (0, 0)

Plug the extreme point

x = \frac{π}{2}

into

3 sin^{2} (x) \Rightarrow y = 3

Maximum (\frac{π}{2}, 3)

Plug the extreme point

x = π

into

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

Minimum (π, 0)

Minimum (0, 0), Maximum (\frac{π}{2}, 3), Minimum (π, 0)

f (x) = 5 x^{3} + 5 x^{2} - 5 x

e^{- x} (2 x + 1) + 3

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

x^{3} + \frac{48}{x}

f (x) = 3 x^{5} - 6 x^{3}

What is the extreme f(x)=3sin^2(x),0<= x<= pi ?
The extreme f(x)=3sin^2(x),0<= x<= pi is Minimum(0,0),Maximum(pi/2 ,3),Minimum(pi,0)