# Difference between revisions of "2004 AMC 12B Problems/Problem 13"

## Problem

If $f(x) = ax+b$ and $f^{-1}(x) = bx+a$ with $a$ and $b$ real, what is the value of $a+b$?

$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 1 \qquad\mathrm{(E)}\ 2$

## Solution (Alcumus)

Since $f(f^{-1}(x))=x$, it follows that $a(bx+a)+b=x$, which implies $abx + a^2 +b = x$. This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$.

Then $b = -a^2$. Substituting into the equation $ab = 1$, we get $-a^3 = 1$. Then $a = -1$, so $b = -1$, and$$f(x)=-x-1.$$Likewise$$f^{-1}(x)=-x-1.$$These are inverses to one another since$$f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.$$$$f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.$$Therefore $a+b=\boxed{-2}$.