(1)y′=-2x+2=2(-x+1)=0;
x=1;
x>1;y′<0;单调递减;
x<1,y′>0;单调递增;
所以极大值f(1)=-1+2+3=4;
(2)y′=1-(3/2)x^(1/2)=0;
x^(1/2)=2/3;
x=4/9;
1-(3/2)x^(1/2)>0;x
(1)y′=-2x+2=2(-x+1)=0;
x=1;
x>1;y′<0;单调递减;
x<1,y′>0;单调递增;
所以极大值f(1)=-1+2+3=4;
(2)y′=1-(3/2)x^(1/2)=0;
x^(1/2)=2/3;
x=4/9;
1-(3/2)x^(1/2)>0;x