Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by \[\mathrm{f} : x \to \frac{4}{(x-4)^2}, \

Question

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\mathrm{f} : x \to \frac{4}{(x-4)^2}, \quad x \in \mathbb{R}, x \neq 4$$$$\mathrm{g} : x \to \ln\left(1 + \frac{1}{x}\right), \quad x \in \mathbb{R}, x > 0$$ Sketch the graph of $y = \mathrm{f}(x)$, stating the equations of any asymptotes, the coordinates of the points where it crosses the axes and the coordinates of the turning points, if any.[2] Show that $\mathrm{g}\mathrm{f}$ exists. Hence find the rule, domain and range of $\mathrm{g}\mathrm{f}$.[4] If the domain of $\mathrm{f}$ is further restricted to $x < k$, state the largest value of $k$ for which the function $\mathrm{f}^{-1}$ exists.[1] Using the restricted domain found in part (c), find $\mathrm{f}^{-1}$ in a similar form.[3]

Accepted Answer

(a) Asymptotes $x=4$, $y=0$; intercept $\left(0,\frac14\right)$; no turning point; sketch shown. (b) $\mathrm{gf}(x)=\ln\left(1+\frac{(x-4)^2}{4}\right)$, domain $\mathbb R\setminus\{4\}$, range $(0,\infty)$. (c) $k=4$. (d) $\mathrm{f}^{-1}:x\mapsto4-\frac2{\sqrt x}$, $x>0$.

2025 CJC P2 Q2

2025 CJC P2 Q2

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