The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \ln \left( {{x}^{2}}-4 \right)$ for

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \ln \left( {{x}^{2}}-4 \right)$ for $2 < x\le 8$. Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.[3] Sketch the graphs of $\mathrm{f}$, ${{\mathrm{f}}^{-1}}$ and $\mathrm{f}\,{{\mathrm{f}}^{-1}}$ on the same axes, showing clearly the relationship between the graphs.[3] The region bounded by the curve $y=\mathrm{f}\left( x \right)$, the line $x=8$ and the $x$-axis is rotated through $2\pi $ radians about the $y$- axis. Find the exact volume of the solid obtained.[4]

Accepted Answer

(i) $\mathrm f^{-1}(x)=\sqrt{e^x+4}$, with domain $(-\infty,\ln60]$ (ii) $y=\mathrm f(x)$ has vertical asymptote $x=2$, intercept $(\sqrt5,0)$, and endpoint $(8,\ln60)$; $y=\mathrm f^{-1}(x)$ has horizontal asymptote $y=2$, intercept $(0,\sqrt5)$, and endpoint $(\ln60,8)$; $\mathrm f\mathrm f^{-1}(x)=x$ for $x\leq\ln60$ (iii) $\pi(60\ln60-59)\text{ units}^3$

2022 TJC Promo Q9

2022 TJC Promo Q9

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