The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{x}^{2}}-4x-5$, for \(x\in \m

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{x}^{2}}-4x-5$, for $x\in \mathbb{R}$, $x\le 2$. Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.[3] On the same diagram, sketch the graphs of $\mathrm{f}$ and ${{\mathrm{f}}^{-1}}$.[3] Find the exact solution of the equation $\mathrm{f}\left( x \right)={{\mathrm{f}}^{-1}}\left( x \right)$.[3] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto 1+{{\mathrm{e}}^{2x}}$, for $x\in \mathbb{R}$, $x<0$. Explain why the composite function $\mathrm{fg}$ exists and find the range of $\mathrm{fg}$.[3]

Accepted Answer

(a) $\mathrm f^{-1}(x)=2-\sqrt{x+9}$, with domain $[-9,\infty)$ (b) For $y=\mathrm f(x)$: endpoint $(2,-9)$, $x$-intercept $(-1,0)$, $y$-intercept $(0,-5)$; for $y=\mathrm f^{-1}(x)$: endpoint $(-9,2)$, $x$-intercept $(-5,0)$, $y$-intercept $(0,-1)$ (c) $x=\dfrac{5-3\sqrt5}{2}$ (d) $\mathrm{fg}$ exists because $R_{\mathrm g}=(1,2)\subset D_{\mathrm f}=(-\infty,2]$, and $R_{\mathrm{fg}}=(-9,-8)$

2022 YIJC Promo Q9

2022 YIJC Promo Q9

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