Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $f$ :/

thanks

share|improve this question
    
See math.stackexchange.com/questions/2349/… –  anon Oct 11 '10 at 21:44

3 Answers 3

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

share|improve this answer

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

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

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

share|improve this answer

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

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.