Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by \(\mathrm{f}:x\mapsto {{\left( x-2 \right

Question

Functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $\mathrm{f}:x\mapsto {{\left( x-2 \right)}^{2}}-1$, $x\in \mathbb{R}$, $x\le 1$, $\mathrm{g}:x\mapsto \frac{1}{1-x}$, $x\in \mathbb{R}$, $x>1$ Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.[3] On the same diagram, sketch the graphs of $y=\mathrm{f}\left( x \right)$ and $y={{\mathrm{f}}^{-1}}\left( x \right)$, showing clearly the relationship between the graphs and giving the coordinates of the axial intercepts. Hence find the exact solution of the equation $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.[5] Determine whether the composite function $\mathrm{fg}$ exists, justifying your answer.[2]

Accepted Answer

(a) $\mathrm f^{-1}(x)=2-\sqrt{x+1}$, with domain $[0,\infty)$ (b) $y=\mathrm f(x)$ has intercepts $(1,0)$ and $(0,3)$, $y=\mathrm f^{-1}(x)$ has intercepts $(3,0)$ and $(0,1)$, and $x=\dfrac{5-\sqrt{13}}{2}$ (c) $\mathrm{fg}$ exists because $R_{\mathrm g}=(-\infty,0)\subset D_{\mathrm f}=(-\infty,1]$

2023 TJC Promo Q8

2023 TJC Promo Q8

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