2. find the absolute maximum and minimum of the function ( f(x)=sin ^{2}x+sin x ) on the interval ( 0,2pi ).

Explanation:

Step1: Let ( t = \sin x )

Since ( x\in[0,2\pi] ), then ( t\in[- 1,1] ). The function ( f(x)=\sin^{2}x+\sin x ) can be rewritten as ( y = t^{2}+t ).

Step2: Find the derivative of ( y = t^{2}+t )

The derivative ( y^\prime=2t + 1 ). Set ( y^\prime = 0 ), then ( 2t+1=0 ), solving for ( t ) gives ( t=-\frac{1}{2} ).

Step3: Evaluate ( y = t^{2}+t ) at critical points and endpoints

• When ( t =-\frac{1}{2} ), ( y=\left(-\frac{1}{2}\right)^{2}-\frac{1}{2}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4} ).
• When ( t = 1 ), ( y=1^{2}+1=2 ).
• When ( t=-1 ), ( y=\left(-1\right)^{2}+\left(-1\right)=0 ).

Answer:

The absolute maximum value is ( 2 ) and the absolute minimum value is ( -\frac{1}{4} ).