Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 5 / b / ii / Solution

For $r>0$, solve $f(s)=r$ with $s>0$. Multiplying by the positive denominator gives $s^2+s=1/r$, so the quadratic formula givesThe minus sign gives a negative number and is inadmissible. The plus sign gives a positive number for every $r>0$, and it is the unique positive solution. Hence the inverse function isIndeed $g(r)(g(r)+1)=1/r$, so $f(g(r))=r$. Conversely, for $s>0$, $\sqrt{1+4/f(s)}=\sqrt{(2s+1{)}^2}=2s+1$, giving $g(f(s))=s$. These verify both inverse function identities.