The function $\mathrm{f}$ is defined by \(\mathrm{f}:x\mapsto \ln \left( {{x}^{2}}-6x+10 \right)-5

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \ln \left( {{x}^{2}}-6x+10 \right)-5$ , $x\in \mathbb{R}$, $x\le 3$. Find ${{\mathrm{f}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{f}}^{-1}}$.[3] Sketch on the same diagram the graphs of $y=\mathrm{f}\left( x \right)$ and $y=\mathrm{f}{{\,}^{-1}}\left( x \right)$, giving the coordinates of the axial intercepts.[3] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \frac{1}{x-5}+x-2$, $x\in \mathbb{R}$, $x\ne 5$. Find the range of $\mathrm{g}{{\mathrm{f}}^{-1}}$.[2] Solve the inequality $\mathrm{f}\left( x \right)<\mathrm{g}\left( x \right)$.[2]

Accepted Answer

(i) $\mathrm f^{-1}(x)=3-\sqrt{e^{x+5}-1}$, with domain $[-5,\infty)$ (ii) For $y=\mathrm f(x)$: $x$-intercept $\left(3-\sqrt{e^5-1},0\right)$, $y$-intercept $(0,\ln 10-5)$, endpoint $(3,-5)$; for $y=\mathrm f^{-1}(x)$: $x$-intercept $(\ln 10-5,0)$, $y$-intercept $\left(0,3-\sqrt{e^5-1}\right)$, endpoint $(-5,3)$ (iii) $R_{\mathrm g\mathrm f^{-1}}=\left(-\infty,\dfrac12\right]$ (iv) $-0.323<x\le 3$, to three significant figures

2022 RI Promo Q10

2022 RI Promo Q10

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