f(x)=(2x+1)/(x-1) with domain x>3. (a)Find the inverse of f(x). (b)Find the range of f(x). (c) g(x)=x+5 for all x. Find the value of x such that fg(x)=3.

(a) Let y=f(x). Then y = (2x+1)/(x-1). Rearrange the equation to get x in terms of y to obtain the inverse function. This gives x=(1+y)/(y-2). So the inverse of f is f^-1(x)=(1+x)/(x-2) (b) Drawing a graph of f(x) gives a vertical asymptote at x=1 and a horizontal asymptote at y=2. This is because for large values of x, f(x) tends to 2x/x = 2. For values of x>1, the graph shows that f(x)>2. Note that the domain of f(x) is x>3, and f(3) = 3.5. So the range of this function is in fact restricted to 2 (c) Recall from part (a) that f^-1(x)=(1+x)/(x-2). By taking f^-1 on both sides of the equation fg(x)=3, we get  f^-1fg(x)=f^-1(3). Note that  f^-1f(x)=x so this gives g(x)= f^-1(3)=4. So g(x)=x+5=4 giving x= -1.