The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by \[\begin{aligned} &\mathrm{f} : x \ma

Question

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\begin{aligned} &\mathrm{f} : x \mapsto 5 - (x - 2)^2, \quad \text{for}\hspace{0.5em} x \in \mathbb{R}, \ x \ge 0, \ &\mathrm{g} : x \mapsto \mathrm{e}^x + 3, \qquad \text{for}\hspace{0.5em} x \in \mathbb{R}. \end{aligned}$$ Show that the composite function $\mathrm{fg}$ exists.[2] Find an expression for $\mathrm{fg}(x)$ and state its domain.[2] Find the range of $\mathrm{fg}$.[2] Show that $\mathrm{f}^{-1}$ does not exist.[1] If the domain of $\mathrm{f}$ is restricted to $[k, \infty)$, find the smallest value of $k$ such that $\mathrm{f}^{-1}$ exists.[1] For the rest of the question, the domain of $\mathrm{f}$ is now restricted to $[k, \infty)$, with the value of $k$ found in part (v). Find $\mathrm{f}^{-1}(x)$ and state the domain of $\mathrm{f}^{-1}$.[3]

Accepted Answer

(i) $\operatorname{range}(g)=(3,\infty)\subseteq[0,\infty)=\operatorname{domain}(f)$, so $fg$ exists (ii) $fg(x)=5-(e^x+1)^2$, with domain $\mathbb R$ (iii) $\operatorname{range}(fg)=(-\infty,4)$ (iv) $f^{-1}$ does not exist because $f$ is not one-to-one (v) $k=2$ (vi) $f^{-1}(x)=2+\sqrt{5-x}$, with domain $(-\infty,5]$

2024 MI PRU1 Q7

2024 MI PRU1 Q7

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