Shifting and inverse functions

[Lord Anoobis]

Homework Statement

If we shift a curve to the left, what happens to its reflection in the line y = x? In view of this geometric principle, find an expression for the inverse of g(x) = f(x + c) where f is a one-to-one function.

Homework Equations

The Attempt at a Solution

Initially I did this, thinking the reflection shifts to the right:


g(x) = f(x + c)
f^-1(g(x)) = x + c
x = f^-1(g(x)) - c
then g-1(x) = f-1(g(x)) - c

I soon realized that the reflection actually shifts downward and the correct answer is slightly different, with the above calculation being` unnecessary. What I would like to know is, where did the above method using the cancellation equations go wrong?

 

[RUber]

When you have $x = f^{-1}(g(x))-c$ the inverse function is what you get when you replace x with $g^{-1}(x)$ .
In so doing you get $g^{-1}(x)= f^{-1}(g(g^{-1}(x)))-c=f^{-1}(x) - c$
f(x+c) is the form of a left shift of c units. When reflected about y=x, you are sending (x,y) to (y, x).

Consider f(x) = x^2 on x>0 , g(x) = (x+2)^2 on x>-2.
$f^{-1}(x) = \sqrt{x}, g^{-1}(x) = \sqrt{x} - 2 = f^{-1} (x)-2$

 

[Lord Anoobis]

When you have $x = f^{-1}(g(x))-c$ the inverse function is what you get when you replace x with $g^{-1}(x)$ .
In so doing you get $g^{-1}(x)= f^{-1}(g(g^{-1}(x)))-c=f^{-1}(x) - c$
f(x+c) is the form of a left shift of c units. When reflected about y=x, you are sending (x,y) to (y, x).

Consider f(x) = x^2 on x>0 , g(x) = (x+2)^2 on x>-2.
$f^{-1}(x) = \sqrt{x}, g^{-1}(x) = \sqrt{x} - 2 = f^{-1} (x)-2$

Thank you for clearing that up.