若函数y=f(x)的定义域为(1,9)且f(x)=lg(x+1),则函数y=f(x）+f(2x)的值域

y = f(x) + f(2x)
y = log(x + 1) + log(2x + 1)
y = log(2x^2 + 3x + 1)
代入x = 1和x = 9,得
y = log(6)和log(190)
因为2x^2 + 3x + 1随x增加单调增加,则y的值域为(log(6),log(190)).