<p>The function $\mathrm{f}$, with domain the set of positive real numbers, is given by</p>
<p style="text-align: center">${\mathrm{f}(x) = \begin{cases}2x + 1 & \text{for } 0 < x < 2 \\ (x - 3)^2 + 6 & \text{for } 2 \leq x \leq 3 \\ 1 & \text{for } x > 3 \end{cases}}$</p>
<ol class="i-list"><li><div class="list-item-paragraph">State whether ${{\mathrm{f}}^{-1}}$ exists, justifying your answer.<span class="inline-marks">[1]</span></div></li></ol>
<p>The function $\mathrm{g}$ is defined by</p>
<p style="text-align: center">$\mathrm{g}:x\mapsto \mathrm{f}\left( x \right)$, $0 < x\le 3$</p>
<ol class="i-list" start="2"><li><div class="list-item-paragraph">Sketch the graph of $y=\mathrm{g}\left( x \right)$, and find the range of $\mathrm{g}$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">State with reasons whether ${{\mathrm{g}}^{2}}$ exist.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">On a separate diagram, sketch the graphs of $y={{\mathrm{g}}^{-1}}\left( x \right)$ and $y={{\mathrm{g}}^{-1}}\mathrm{g}\left( x \right)$.<span class="inline-marks">[3]</span></div></li></ol>

The function $\mathrm{f}$, with domain the set of positive real numbers, is given by

${\mathrm{f}(x) = \begin{cases}2x + 1 & \text{for } 0 < x < 2 \\ (x - 3)^2 + 6 & \text{for } 2 \leq x \leq 3 \\ 1 & \text{for } x > 3 \end{cases}}$

State whether ${{\mathrm{f}}^{-1}}$ exists, justifying your answer.

$\mathrm{g}:x\mapsto \mathrm{f}\left( x \right)$, $0 < x\le 3$

Sketch the graph of $y=\mathrm{g}\left( x \right)$, and find the range of $\mathrm{g}$.

State with reasons whether ${{\mathrm{g}}^{2}}$ exist.

On a separate diagram, sketch the graphs of $y={{\mathrm{g}}^{-1}}\left( x \right)$ and $y={{\mathrm{g}}^{-1}}\mathrm{g}\left( x \right)$.

2019 VJC J1 MYE Q9

2019 VJC J1 MYE Q9

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