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Can you help me understand how

$$g^{-1}(x) = f^{-1}(x + 2)$$

is the inverse of

$$g(x) = f(x) - 2$$

I cannot get my head around how the 2 ends up inside $f$ :/


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See… – anon Oct 11 '10 at 21:44
up vote 5 down vote accepted

If you want to complement Bill Dubuque's answer with trying to figure out how you might have come up with the formula, think about what $g(x)=f(x)-2$ is telling you to do: in order to compute $g(x)$, you first compute $f(x)$, and then you subtract $2$. The inverse of this should be something that "unravels" the process you just went to. So the first thing you need to unravel is the last thing you did when you computed $g(x)$, namely subtracting $2$. So the first thing you are going to do is add $2$ from whatever they give you. What did you did just prior to subtracting $2$ when you computed $g$? You computed $f$. So the next thing you need to do in the "unraveling" process is to undo the computation of $f$, which is achieved by computing $f^{-1}$.

In summary: to invert (unravel) the process of (i) first compute $f$, then (ii) subtract $2$; you will (I) first add $2$; then (II) compute $f^{-1}$. Symbolically: if you are given $x$ and you want to unravel, first you compute $x+2$, then you compute $f^{-1}$ of that. So if $g(x) = f(x)-2$, then $g^{-1}(x) = f^{-1}(x+2)$.

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It's simply $\rm\ g = h\: f\ \Rightarrow\ g^{-1} = (h\:f)^{-1} = f^{-1} h^{-1},\ \ h(x) = x-2\:.\ $

Or you can explicitly check that they're inverses, namely

$\rm\ g^{-1}(g(x)) = f^{-1}(g(x)+2) = f^{-1}(f(x)-2 + 2) = x$

$\rm\ g(g^{-1}(x)) = f(g^{-1}(x))-2 = f(f^{-1}(x+2))-2 = x$

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$$g(x) = f(x) - 2 = (h \circ f)(x)$$

where $h(x) = x - 2$ and of course $h^\circ(x) = x + 2$

Then the inverse $g^{\circ}(x) = (h \circ f)^{\circ}(x) = (f^{\circ}\circ h^{\circ})(x) = f^\circ(x + 2)$ by the usual rules.

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