Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 8

The domain of a function excludes $x=3$. Both branches are positive, with a vertical asymptote at $x=3$ approached from both sides with value $+\infty$, and horizontal asymptote $y=0$ at either infinity. The graph is symmetric about $x=3$ because it depends only on $|x-3|$.

For $x<3$, the formula is $2/(3-x)$ and the derivative is positive; for $x>3$, it is $2/(x-3)$ and the derivative is negative. It passes through $(0,2/3)$ and has no zeros.

The sine addition formula gives $(\sin x+\cos x)/\sqrt{2}=\sin (x+\pi /4)$. On the requested interval the shifted angle runs from $0$ to $\pi /2$, so the graph rises from $(-\pi /4,0)$ through $(0,1/\sqrt{2})$ to $(\pi /4,1)$.

Its derivative $\cos (x+\pi /4)$ is nonnegative and reaches zero only at the right endpoint. Its second derivative is $-\sin (x+\pi /4)\le 0$, so the rising arc is concave downward and ends with a horizontal tangent.