First I'll teach you a little bit about inverse functions. Stuff you should know before you try to do inverse problems:This is the graph of the inverse of f(x), which we call f-1(x). The inverse of a function is the graph made by interchanging the x-coordinates with the y-coordinates of the points on the graph. That is to say that if (a,b) is a point on f(x), then (b,a) is a point on f-1(x). If the equation for f-1(x) were simpler we could calculate it and graph it. But we can't do that here because we don't know how to solve a 5th degree polynomial for x. I will find a few points on f(x), by finding a few points on f-1(x) and reversing the coordinates. x f-1(x) point on f-1(x) point on f(x) ------------------------------------------------ -1 -5.00 (-1,-5.00) (-5.00,1) -0.9 -3.75 (-0.9,-3.75) (-3.75,-0.9) -0.8 -2.75 (-0.8,-2.75) (-2.75,-0.8) -0.7 -1.95 (-0.7,-1.95) (-1.95,-0.7) -0.5 -0.78 (-0.5,-0.78) (-0.78,-0.5) 0.0 1.00 (0,1.00) (1.00,0) 0.3 1.96 (0.3,1.96) (1.96,0.3) 0.5 2.78 (0.5,2.78) (2.78,0.5) 0.7 3.95 (0.7,3.95) (3.95,0.7) 0.8 4.75 (0.8,4.75) (4.75,0.8) Here are those 10 points of f(x) plotted: Now we'll look at your questions: (a) Compute f-1(1) and f(1) To find f-1(1), substitute x=1 in So f-1(1) = 7 That means the point (1,7) is on the graph of f-1(x). We want to find f(1). We notice that since ends with a 1. Then if x=0 then So the point (0,1) is on the graph of f-1(x), so the point (1,0) is on the graph of f(x), so f(1) = 0 (b) Compute the value of such that . That means (x0,1) must be on f(x), so its reverse, (1,x0) must be on f-1(x), and since f-1(1) = 7, (1,7) is on f-1(x) and that means (7,1) is on f(x), so therefore f(7)=1 so x0 = 7 (c) Compute the value of such that . That means (y0,1) must be on f-1(x), so its reverse, (1,y0) must be on f(x), and since f(1) = 0, (1,0) is on f(x) and that means (0,1) is on f-1(x), so therefore f-1(0)=1 so y0 = 0. Edwin