The function $\mathrm{f}(x)$ is defined by $\mathrm{f}:x\mapsto\mathrm{e}^{\left|3-x\right|}$, \

Question

The function $\mathrm{f}(x)$ is defined by $\mathrm{f}:x\mapsto\mathrm{e}^{\left|3-x\right|}$, $x\in\mathbb{R}$, $x\leq m$. Explain why $3$ is the greatest value of $m$ for which the inverse function $\mathrm{f}^{-1}$ exists.[1] For the remainder of the question, $m=3$. Find $\mathrm{f}^{-1}(x)$ and state its domain.[3] Solve the inequality $\mathrm{f}\left(\mathrm{f}^{-1}(\left|x\right|)\right)\geq 1$.[3] The function $\mathrm{g}(x)$ is defined by $\mathrm{g}:x\mapsto n+\sqrt{x-4}$, $x\in\mathbb{R}$, $x\geq 4$ and $n\in\mathbb{R}$. Find the smallest value of $n$ for the composite function $\mathrm{f}^{-1}\mathrm{g}(x)$ to exist. Justify your answer.[2]

Accepted Answer

(a) For $x\leq 3$, $\mathrm{f}(x)=\mathrm{e}^{3-x}$ is strictly decreasing, so $\mathrm{f}$ is one-one. For $m>3$, $\mathrm{f}(3-t)=\mathrm{f}(3+t)$ for $0 < t\leq m-3$, so $\mathrm{f}$ is not one-one (b) $\mathrm{f}^{-1}(x)=3-\ln x$, with domain $x\geq 1$ (c) $x\leq -1$ or $x\geq 1$ (d) $n=1$, because $\mathrm{f}^{-1}\mathrm{g}$ needs $R_{\mathrm{g}}=[n,\infty)\subseteq D_{\mathrm{f}^{-1}}=[1,\infty)$

N2026 P1 Q10

N2026 P1 Q10