If f(x) = (3x-2) / x-5 x>6, find a.) ff(8) b.) the range of f(x) c.) f^-1(x) and state its range.

Firstly, ff(8) is the same as f(f(8)) so f(8) needs to be found first. Subbing in x=8 f(8)=22/3, so f(22/3) is required, Setting x=22/3 f(22/3)= 20/(22/3 - 5) = 60/7. so ff(8)=60/7.Next, as x tends to infinity f(x) tends to 3 at x=6 f(x) = 16. The function never reaches either limit however so 3<f(x)<16.Finally for c.), set y=f(x). The inverse function is a reflection of the original function in f(x), so rearrange y = (3x-2)/x-5 to get x as a function of y. Multiplying both sides by x-5, yx-5y = 3x-2 next group all terms with x to get 3x-yx=2-5y. x(3-y) = 2-5y, so x = (2-5y)/(3-y). Now swap the x and y's, which is the equivalent or the reflection in the line y=x. y = (2-5x)/(3-x) so f^-1(x) = 2-5x/3-x. The range of the inverse of a function is the domain of the original function, so f^-1(x)>6.