A function $\mathrm{h}$ is self-inverse if $\mathrm{h}(x) = \mathrm{h}^{-1}(x)$ for all \(x

Question

A function $\mathrm{h}$ is self-inverse if $\mathrm{h}(x) = \mathrm{h}^{-1}(x)$ for all $x$ in the domain of $\mathrm{h}$. The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\mathrm{f}(x) = 2 + \frac{1}{x-2}, \quad \text{for}\hspace{0.5em} x \in \mathbb{R}, x > 2$$ $$\mathrm{g}(x) = \begin{cases} 3x - 1 & \text{for}\hspace{0.5em} x \le 1, \ (x - 1)^2 + 2 & \text{for}\hspace{0.5em} x > 1. \end{cases}$$ Sketch the graph of $y = \mathrm{f}(x)$. With the aid of your graph, explain why $\mathrm{f}$ has an inverse.[2] Show that $\mathrm{f}$ is self-inverse and find $\mathrm{f}^2(x)$.[4] Hence, or otherwise, evaluate $\mathrm{f}^{2025}(4)$.[2] Find an expression for $\mathrm{g}\mathrm{f}$ and state its domain.[3] Find the range of $\mathrm{g}\mathrm{f}$.[2]

Accepted Answer

(a) $\mathrm{f}$ is one-to-one, so $\mathrm{f}^{-1}$ exists (b) $\mathrm{f}^{-1}(x)=\mathrm{f}(x)$, $\mathrm{f}^2(x)=x$ for $x>2$ (c) $\displaystyle \mathrm{f}^{2025}(4)=\frac{5}{2}$ (d) $\displaystyle \mathrm{g}\mathrm{f}(x)=\left(\frac{x-1}{x-2}\right)^2+2$, domain $(2,\infty)$ (e) range $(3,\infty)$

2024 CJC Promo Q11

2024 CJC Promo Q11

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