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

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

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

extreme

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

Decreasing

: 0 < x < \frac{π}{2},

Increasing

: \frac{π}{2} < x < π

Plug the extreme point

x = 0

into

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

Maximum (0, 0)

Plug the extreme point

x = \frac{π}{2}

into

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

Minimum (\frac{π}{2}, - 3)

Plug the extreme point

x = π

into

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

Maximum (π, 0)

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

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