<p>The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{x-1}{x}$, $x\in \mathbb{R}$, $x\ne 0$ and $x\ne 1$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Define ${{\mathrm{f}}^{-1}}$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Show that the composite function ${{\mathrm{f}}^{2}}$ exists and that&nbsp;${{\mathrm{f}}^{2}}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">State ${{\mathrm{f}}^{3}}\left( x \right)$.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;${{\mathrm{f}}^{2014}}\mathrm{g}\left( x \right)=1-{{\mathrm{e}}^{-x}}$ for&nbsp;$x&gt;0$, find $\mathrm{g}\left( x \right)$.<span class="inline-marks">[3]</span></div></li></ol>

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{x-1}{x}$, $x\in \mathbb{R}$, $x\ne 0$ and $x\ne 1$.

Show that the composite function ${{\mathrm{f}}^{2}}$ exists and that ${{\mathrm{f}}^{2}}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.

State ${{\mathrm{f}}^{3}}\left( x \right)$.

Given that ${{\mathrm{f}}^{2014}}\mathrm{g}\left( x \right)=1-{{\mathrm{e}}^{-x}}$ for $x>0$, find $\mathrm{g}\left( x \right)$.

2014 JJC Promo Q8

2014 JJC Promo Q8

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