The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto 3+2x-{{x}^{2}}$, \(x\in \mathbb{R}\

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto 3+2x-{{x}^{2}}$, $x\in \mathbb{R}$, $x>1$ Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.[3] On the same axes, sketch the graphs of $y=\mathrm{f}\left( x \right)$, $y={{\mathrm{f}}^{-1}}\left( x \right)$ and $y={{\mathrm{f}}^{-1}}\mathrm{f}\left( x \right)$.[3] Find the exact value of $x$ that satisfies the equation $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.[2] Another function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \sqrt{4-x}$, $x\in \mathbb{R}$, $x\le 4$ Explain why the composite function $\mathrm{gf}$ exists and find $\mathrm{gf}\left( x \right)$, simplifying your answer.[3]

Accepted Answer

(a) $\mathrm f^{-1}(x)=1+\sqrt{4-x}$, with domain $(-\infty,4)$ (b) $y=\mathrm f(x)$ has open endpoint $(1,4)$ and $x$-intercept $(3,0)$; $y=\mathrm f^{-1}(x)$ has open endpoint $(4,1)$ and $y$-intercept $(0,3)$; $y=\mathrm f^{-1}\mathrm f(x)=x$ for $x>1$ (c) $x=\dfrac{1+\sqrt{13}}{2}$ (d) $\mathrm{gf}$ exists because $R_{\mathrm f}=(-\infty,4)\subset D_{\mathrm g}=(-\infty,4]$, and $\mathrm{gf}(x)=x-1$ for $x>1$

2023 YIJC Promo Q9

2023 YIJC Promo Q9

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