[1]
∵2x²≦3x,即x[x-(3/2)]≦0
∴0≦x≦3/2
即x∈[0,3/2]
[2]
函数f(x)=[x+(1/2)]²+(3/4)
∴当0≦x≦3/2时
有f(0)≦f(x)≦f(3/2)
即此时f(x)min=f(0)=1,
f(x)max=f(3/2)=19/4