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

The extreme f(x)=sin^2(x),0<= x<= (2pi)/3 is Minimum(0,0),Maximum(pi/2 ,1),Minimum((2pi)/3 , 3/4)

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

extreme

Minimum (0, 0), Maximum (\frac{π}{2}, 1), Minimum (\frac{2 π}{3}, \frac{3}{4})

Solution steps

Find the critical points:

x = 0, x = \frac{π}{2}, x = \frac{2 π}{3}

Domain of

sin^{2} (x), 0 \leq x \leq \frac{2 π}{3} : 0 \leq x \leq \frac{2 π}{3}

Find intervals:

Increasing

: 0 < x < \frac{π}{2},

Decreasing

: \frac{π}{2} < x < \frac{2 π}{3}

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 = \frac{2 π}{3}

into

sin^{2} (x) \Rightarrow y = \frac{3}{4}

Minimum (\frac{2 π}{3}, \frac{3}{4})

Minimum (0, 0), Maximum (\frac{π}{2}, 1), Minimum (\frac{2 π}{3}, \frac{3}{4})

What is the extreme f(x)=sin^2(x),0<= x<= (2pi)/3 ?
The extreme f(x)=sin^2(x),0<= x<= (2pi)/3 is Minimum(0,0),Maximum(pi/2 ,1),Minimum((2pi)/3 , 3/4)