The functions ${\mathrm{f}}$ and ${\mathrm{g}}$ are defined by \({\mathrm{f}:x\mapsto \sqrt{x-2}

Question

The functions ${\mathrm{f}}$ and ${\mathrm{g}}$ are defined by ${\mathrm{f}:x\mapsto \sqrt{x-2}}$, ${x\in \mathbb{R}}$, ${x\ge 2}$ ${\mathrm{g}:x\mapsto 3{{x}^{2}}-6x+5}$, ${x\in \mathbb{R}}$, ${0\le x\le 4}$. Sketch the graph of $y=\mathrm{f}\left( x \right)$ and state the range of $\mathrm{f}$.[2] Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and write down the domain of ${{\mathrm{f}}^{-1}}$.[2] Show that the composite function ${\mathrm{fg}}$ exists.[2] Find $\mathrm{fg}\left( x \right)$ and find the range of ${\mathrm{fg}}$.[3] The domain of $\mathrm{g}$ is now defined to be $x\in \mathbb{R}$. Describe fully a sequence of transformations which transforms the curve $y=\mathrm{g}\left( x \right)$ onto the curve ${y={{x}^{2}}-1}$.[3]

Accepted Answer

(i) $~{{\text{R}}_{\text{f}}}=\left[ 0,\infty \right)$ (ii) Domain of ${{\text{f}}^{-1}}:\left[ 0,\infty \right)$ (iii) Since ${{\text{R}}_{\text{g}}}=\left[ 2,29 \right]\subseteq {{\text{D}}_{\text{f}}}=\left[ 2,\infty \right)$. $\text{fg}$ exists. (iv) ${{\text{R}}_{\text{fg}}}=\left[ 0,\sqrt{27} \right]$ (v) 1. Translate in negative $x$-direction by $1$ unit. 2. Translate in negative $y$-direction by $5$ units. 3. Scale parallel to $y$-axis by scale factor $\frac{1}{3}$

2023 MI PRU2 P1 Q8

2023 MI PRU2 P1 Q8

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