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

The domain of a function is fixed by its declared type, soAlthough the rational expression also has values at some negative arguments, they are not in the specified domain of a function. Part (a) of the printed question only supplies the definition used in these four subparts.^[ref]

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.

For $s>0$, the denominator $s(s+1)$ is positive, so every value of $f$ is positive. Conversely, for every $r>0$, the positive number $s=(\sqrt{1+4/r}-1)/2$ satisfies $f(s)=r$. Thus its image of a function, or range, isOne can also see the full range from continuity and monotonicity: $f$ is a continuous function with $f^{\prime }(s)=-(2s+1)/(s^2+s{)}^2<0$, and its endpoint limits are $+\infty$ as $s\to 0^{+}$ and $0$ as $s\to \infty$. The endpoints themselves are not attained.

The partial fraction decompositionmakes the sum a telescoping sum. All interior terms cancel: