Functions: ACJC 2025 Promo Q8

(a)

Writing $\mathrm{f}(x) = \dfrac{ax}{x-a} = a + \dfrac{a^2}{x-a}$, the asymptotes are $x = a$ and $y = a$, and the endpoint is $\left(-a, \tfrac{a}{2}\right)$.

Every horizontal line $y = k$ meets the graph at most once, so $\mathrm{f}$ is one-one and $\mathrm{f}^{-1}$ exists.

(b)

Let $y = \dfrac{ax}{x-a}$: $$\begin{aligned} xy - ay &= ax\\ x(y - a) &= ay\\ x &= \frac{ay}{y - a}. \end{aligned}$$ Hence $\mathrm{f}^{-1}(x) = \dfrac{ax}{x - a}$, with domain $\left(-\infty,\, \dfrac{a}{2}\right]$.

(c)

Since $\mathrm{f}^{-1}$ has the same rule as $\mathrm{f}$, the curves $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ coincide on the intersection of their domains: $$\begin{aligned} -a \leq x \leq \frac{a}{2}. \end{aligned}$$

(d)

As $\mathrm{f}^2(x) = x$, we have $\mathrm{f}^{2k}(x) = x$ and $\mathrm{f}^{2k+1}(x) = \mathrm{f}(x)$. Therefore $$\begin{aligned} \mathrm{f}^{2k+1}\!\left(-\frac{1}{2}a\right) &= \mathrm{f}\!\left(-\frac{1}{2}a\right) = \frac{a\left(-\tfrac{1}{2}a\right)}{-\tfrac{1}{2}a - a} = \frac{-\tfrac{1}{2}a^2}{-\tfrac{3}{2}a} = \frac{a}{3}. \end{aligned}$$

(e)

From $\mathrm{fg}(x) = x + a$: $$\begin{aligned} \mathrm{g}(x) &= \mathrm{f}^{-1}(x + a) = \frac{a(x+a)}{(x+a) - a} = \frac{a(x+a)}{x} = a + \frac{a^2}{x}, \qquad x \leq -a. \end{aligned}$$