设函数f(x)与g(x)分别满足f(x+1/2)=log1/2 (x²-9/4)和g(x)=log1/2 (x-1)-1
问题描述:
设函数f(x)与g(x)分别满足f(x+1/2)=log1/2 (x²-9/4)和g(x)=log1/2 (x-1)-1
(1)求函数f(x)的表达式;
(2)若f(x)>g(x),求x的取值范围.
答
(1) 令t=x+1/2,则x=t-1/2 f(t)=log1/2[(t-1/2)^2-9/4]=log1/2(t^2-t-2)
故 f(x)=log1/2(x^2-x-2)
(2) f(x)=log1/2(x^2-x-2)>g(x)=log1/2(x-1)-1=log1/2(x-1)-log1/2(1/2)=log1/2(2x-2) (1)
1/20; 2x-2>0; x^2-x-2