The function $\mathrm{f}$ is defined by \[\mathrm{f} : x \mapsto \frac{1}{(x-2)^2+1},\quad x \in \

Question

The function $\mathrm{f}$ is defined by $$\mathrm{f} : x \mapsto \frac{1}{(x-2)^2+1},\quad x \in \mathbb{R}.$$ Sketch the graph of $y = \mathrm{f}(x)$, indicating the coordinates of the turning point and the equation of any asymptotes. Explain why $\mathrm{f}^{-1}$ does not exist.[3] If the domain of $\mathrm{f}$ is restricted to $[k, \infty)$, state the smallest value of $k$ for which the function $\mathrm{f}^{-1}$ exists.[1] Using the value of $k$ found in part (ii), find $\mathrm{f}^{-1}(x)$ and state its domain.[3]

Accepted Answer

(i) Maximum point $(2,1)$, asymptote $y=0$, and $f^{-1}$ does not exist on $\mathbb R$ (ii) $k=2$ (iii) $f^{-1}(x)=2+\sqrt{\dfrac1x-1}$, $0<x\le1$

2024 MI PRU2 P2 Q2

2024 MI PRU2 P2 Q2

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