The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto 1-\frac{4}{{{x}^{2}}}$, for \(

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto 1-\frac{4}{{{x}^{2}}}$, for $x\in \mathbb{R}$, $x<-1$. Sketch the graph of $y=\mathrm{f}\left( x \right)$, stating the equation of the asymptote and the coordinates of the point where the curve crosses the axes.[2] Show that $\mathrm{f}$ has an inverse. Hence, find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state its domain.[4] By considering the domains of $y=\mathrm{f}{{\mathrm{f}}^{-1}}\left( x \right)$ and $y={{\mathrm{f}}^{-1}}\mathrm{f}\left( x \right)$, find the range of values of $x$ such that $\mathrm{f}{{\mathrm{f}}^{-1}}\left( x \right)={{\mathrm{f}}^{-1}}\mathrm{f}\left( x \right)$.[2] Another function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto -{{\mathrm{e}}^{2x}}-2$, for $x\in \mathbb{R}$. Explain why $\mathrm{fg}$ exists. Find $\mathrm{fg}$ and its range.[3]

Accepted Answer

(ii) ${{	ext{D}}_{{{\mathrm{f}}^{-1}}}}=\left( -3,1 ight)$ (iii) For $\mathrm{f}{{\mathrm{f}}^{-1}}\left( x ight)={{\mathrm{f}}^{-1}}\mathrm{f}\left( x ight)$, we have $-3<x<-1$. (iv) ${{\mathrm{R}}_{\mathrm{fg}}}=\left( 0,1 ight)$

2024 JPJC Promo Q13

2024 JPJC Promo Q13

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