It is given that \({\mathrm{f}(x) = \begin{cases} \sqrt{4 - (x - 2)^2} & \text{for}\hspace{0.5em} 0

Question

It is given that ${\mathrm{f}(x) = \begin{cases} \sqrt{4 - (x - 2)^2} & \text{for}\hspace{0.5em} 0 < x \leq 2, \ \frac{1}{2}x + c & \text{for}\hspace{0.5em} 2 < x \leq 4, \end{cases}}$ where $c\ge 0$ , and that $\mathrm{f}\left( x \right)=\mathrm{f}\left( x+4 \right)$ for all real values of $x$. State the range of values of $c$ such that the inverse of $\mathrm{f}$ exists.[1] For the minimum value of $c$ obtained in part (i), sketch the graph of $y=\mathrm{f}\left( x \right)$ for $-4\le x\le 5$.[3] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \frac{1}{x}$ for $x>0$. State the range of $\mathrm{g}$. Hence show that the composite function ${{\mathrm{g}}^{-1}}\mathrm{f}$ exists.[2] Solve the equation ${{\mathrm{g}}^{-1}}\mathrm{f}\left( x \right)=1$.[3]

Accepted Answer

(i) $c\geq1$ (ii) Period-$4$ sketch with $\mathrm{f}(5)=\sqrt{3}$ (iii) $\mathrm{R}_{\mathrm{g}}=(0,\infty)$; $\mathrm{g}^{-1}\mathrm{f}$ exists (iv) $x=2-\sqrt{3}+4n$, $n\in\mathbb{Z}$

2017 RVHS Promo Q9

2017 RVHS Promo Q9

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