The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 + \frac{3}{\mathrm{e}^x - 2}$

Question

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto 1 + \frac{3}{\mathrm{e}^x - 2}$ for $x \in \mathbb{R}$, $x > \ln 2$. Find $\mathrm{f}^{-1}(x)$ and state its domain.[3] Another function $\mathrm{g}$ is defined by $$\mathrm{g} : x \mapsto \begin{cases} 2(x - 1)^2 + 2 & \text{for}\hspace{0.5em} x \le 0, \ 2 + |2 - x| & \text{for}\hspace{0.5em} 0 < x \le 3. \end{cases}$$ Sketch the graph of $y = \mathrm{g}(x)$ and explain why the composite function $\mathrm{f}^{-1}\mathrm{g}$ exists.[4] Hence find the exact value of $k$ such that $\mathrm{f}^{-1}\mathrm{g}(k) = \ln \frac{5}{2}$.[3]

Accepted Answer

(a) $f^{-1}(x)=\ln\left(\dfrac3{x-1}+2\right)$, $\operatorname{domain}(f^{-1})=(1,\infty)$ (b) graph through $(0,4)$, $(2,2)$, and $(3,3)$ as shown, and $f^{-1}g$ exists (c) $k=1-\dfrac{\sqrt{10}}2$

2025 RI P1 Q7

2025 RI P1 Q7

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