Domain/Range, and Inverse

• Jan 16th 2010, 11:06 PM
goliath
Domain/Range, and Inverse
Hey guys, I can't figure these out,

(refer to the posted graph)

a. Evaluate f^-1 (1).
b. If If g(x) = x^2 + 1, evaluate (f^-1 o g)(1) and (g o f )(1).
c. Graph y = 1 – f(x + 3).
d. State the domain and range of y = 1 – f (x + 3).

I can do part d, if someone could explain why there's an "f" in front of (x + 3) or if someone shows me the graph.
• Jan 16th 2010, 11:55 PM
gyan1010
been a while, I'd wait for someone to confirm my answers or tell me I'm full of it. Anyway...

a) Inverse of f switch x and y values. so you get a curve in the same quadrant with it curving up toward x = infinity. So evaluating x=1 is like finding x where y=1 on the original graph. Which is 0.

b)
(f^-1 o g)(1) => Well g(1) = 2, and f^-1(2) = 1 (remember its like finding x where y=2 in f).
(g o f )(1) => f(1)=2, g(2) = 2^2 + 1 = 5

c) The graph will start at (-3,0), curve down through points (-2,-1) & (-1,-3), ad end at point (0,-7).
To explain...
1st) The f(x+3) is like moving the x-axis 3 to the right. Which has the effect of moving your graph 3 to the left. Making the domain [-3,0] instead of [0,3].
2nd) The f(x+3) is subtracted from something so basically whatever y value comes out of it is made negative. This has the effect of flipping the graph about the x-axis. So at this point it curves down from (-3,-1) to (0,-8). So the range at this point is [-1,-8]
3rd) The previous step was subtracted from 1, so we add 1 to all the y values which moves the graph up 1. So the final range is [-0,-7]

d) see explanations to part c