The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{3x-2}{3x-3}$, $x

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{3x-2}{3x-3}$, $x<1$. Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and the domain of ${{\mathrm{f}}^{-1}}$.[4] Hence or otherwise find ${{\mathrm{f}}^{2}}\left( x \right)$ and state the value of ${{\mathrm{f}}^{2021}}\left( 0.5 \right)$.[3] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \sqrt{9+x}$, $x\in \mathbb{R}$, $0\le x<25$. Sketch the graph of $y=\mathrm{g}\left( x \right)$. Your sketch should indicate the coordinates of the end-points.[2] On the same diagram of part (i), sketch the graph of $y={{\mathrm{g}}^{-1}}\left( x \right)$, showing the relationship between $\mathrm{g}$ and ${{\mathrm{g}}^{-1}}$.[3] State the solution set of $x$ that satisfy the equation ${{\mathrm{g}}^{-1}}\mathrm{g}\left( x \right)=\mathrm{g}{{\mathrm{g}}^{-1}}\left( x \right)$.[1]

Accepted Answer

(a)(i) $\displaystyle \mathrm{f}^{-1}:x\mapsto\frac{3x-2}{3x-3},\ x<1$ (a)(ii) $\mathrm{f}^2(x)=x$ for $x<1$, and $\displaystyle \mathrm{f}^{2021}\left(\frac{1}{2}\right)=\frac{1}{3}$ (b)(i) $y=\sqrt{9+x}$ from $(0,3)$ inclusive to $(25,\sqrt{34})$ exclusive (b)(ii) $y=x^2-9$ from $(3,0)$ inclusive to $(\sqrt{34},25)$ exclusive, reflected in $y=x$ (b)(iii) $\{x\in\mathbb{R}:3\leq x<\sqrt{34}\}$

2020 SAJC Promo Q8

2020 SAJC Promo Q8

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