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

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{x-1}{x}$, $x\in \mathbb{R}$, $x\ne 0$ and $x\ne 1$. Define ${{\mathrm{f}}^{-1}}$.[3] Show that the composite function ${{\mathrm{f}}^{2}}$ exists and that ${{\mathrm{f}}^{2}}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.[3] State ${{\mathrm{f}}^{3}}\left( x \right)$.[1] Given that ${{\mathrm{f}}^{2014}}\mathrm{g}\left( x \right)=1-{{\mathrm{e}}^{-x}}$ for $x>0$, find $\mathrm{g}\left( x \right)$.[3]

Accepted Answer

(i) ${\mathrm{f}^{-1}:x \mapsto \frac{1}{1-x}\,,\; x \in \mathbb{R},\; x 
eq 0 	ext{ and } x 
eq 1}$ (iii) ${\mathrm{f}^3(x)=x}$ (iv) ${\mathrm{g}(x) = \mathrm{e}^x}$

2014 JJC Promo Q8

2014 JJC Promo Q8

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