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 points

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

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})

3 x^{2} e^{x}

f (x) = 4 x^{\frac{1}{3}} + x^{\frac{4}{3}}

f (x) = \frac{1}{6} x^{3} - x^{2} - 6 x + 11

f (x) = \frac{ln ( x )}{5 x}, 1 \leq x \leq 4

f (x) = x^{2} + y^{2} - 6 x + 2

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)