Question

Can someone point me in the right direction, please?
f(x) is one-to-one. What is an expression for the inverse of g(x) = f(x+c)?

Answer 1

@fission-mailed hi i can help you

Answer 2

@Princer_Jones Thank you :) I know that since f(x) is one-to-one, then \[f^{-1}(x)\] exists. But I'm not quite sure where to go from there.

Answer 3

A mapping f is said to be one-to-one if f(a)=f(b) implies a=b.

Answer 4

what is g(x) here?

Answer 5

@fission-mailed

Answer 6

Hold on trying to figure it out...

Answer 7

So 1-1 functions have only one y-value for every x-value. And to find an inverse of a function algebraically you swap the x's and y's and solve for y to get the inverse. So I was thinking y = f(x+c) becomes x = f(y+c), in trying to get the inverse of g(x). But this doesn't seem to make sense, I think I've done something wrong here.

Answer 8

what are you asked to find?? g(x) or g^-1 (x)?

Answer 9

I am supposed to find an expression for the inverse of g(x) = f(x+c)

Answer 10

here g(x) is f(x+c)?

Answer 11

g(x)=f(x+c)???

Answer 12

Yes

Answer 13

\[
g(x) = f(x+c) \\
\text{Let }~h(x) = x + c. ~~ \text{ Then, } \\
g(x) = f(h(x)) \\
g^{-1}(x) = \left( f(h(x)) \right)^{-1} = h^{-1}(f^{-1}(x)) \\
h(x) = x + c \\
h^{-1}(x) = x - c \\
h^{-1}(f^{-1}(x)) = f^{-1}(x) - c \\
g^{-1}(x) = f^{-1}(x) ~-~ c
\]

Answer 14

The important thing to remember here is the formula for the inverse of composite functions. If p(x) and q(x) are two functions, then:
\[
(p \circ q)^{-1}(x) = (q^{-1} \circ p^{-1})(x) \\
\text{ } \\
\text{Another form of the same formula is:}\\
\text{ } \\
\left\{ p(q(x)\right\}^{-1} = q^{-1}(p^{-1}(x))
\]

Answer 15

Thanks for the help. I didn't know the formula for the inverse of composite functions before, but it makes sense now.