(1)a=1时,f(x)=(x-1)/(x+1)
f'(x)=(x+1-(x-1))/(x+1)^2=2/(x+1)^2>0
∴f(x)在定义域[0,3]上为单调递增函数
最小值为f(0)=-1,最大值为f(3)=1/2
(2)f(x)=(ax-1)/(x+1)
f'(x)=(a(x+1)-(ax-1))/(x+1)^2=(a+1)/(x+1)^2
欲使f(x)为减函数,即f'(x)
(1)a=1时,f(x)=(x-1)/(x+1)
f'(x)=(x+1-(x-1))/(x+1)^2=2/(x+1)^2>0
∴f(x)在定义域[0,3]上为单调递增函数
最小值为f(0)=-1,最大值为f(3)=1/2
(2)f(x)=(ax-1)/(x+1)
f'(x)=(a(x+1)-(ax-1))/(x+1)^2=(a+1)/(x+1)^2
欲使f(x)为减函数,即f'(x)