What is the extreme f(x)=5sin^2(x),0<= x<= pi ?

The extreme f(x)=5sin^2(x),0<= x<= pi is Minimum(0,0),Maximum(pi/2 ,5),Minimum(pi,0)

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extreme f(x)=5sin^2(x),0<= x<= pi

extreme points

f (x) = 5 sin^{2} (x), 0 \leq x \leq π

Minimum (0, 0), Maximum (\frac{π}{2}, 5), Minimum (π, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = π

Domain of

5 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

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

Minimum (0, 0)

Plug the extreme point

x = \frac{π}{2}

into

5 sin^{2} (x) \Rightarrow y = 5

Maximum (\frac{π}{2}, 5)

Plug the extreme point

x = π

into

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

Minimum (π, 0)

Minimum (0, 0), Maximum (\frac{π}{2}, 5), Minimum (π, 0)

f (x, y) = x^{3} + y^{3} - 3 x^{2} - 9 y^{2} - 9 x

f (x, y) = \frac{1}{3} 36 - 9 x^{2} - 4 y^{2}

y + v

y = \frac{4 x}{x ^{2} + 1}

f (x) = - 5 x^{6} ln (x), (0, 4)

What is the extreme f(x)=5sin^2(x),0<= x<= pi ?
The extreme f(x)=5sin^2(x),0<= x<= pi is Minimum(0,0),Maximum(pi/2 ,5),Minimum(pi,0)