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

Question

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \frac{1}{(x+2)^2}$, $x \in \mathbb{R}$, $x \neq -2$. The domain of $\mathrm{f}$ is further restricted to be $x > a$, where $a$ is an integer. State the least value of $a$ such that the function $\mathrm{f}^{-1}$ exists.[1] Hence find $\mathrm{f}^{-1}(x)$ and state its domain.[3] Sketch the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{f}^{-1}(x)$ on the same diagram, giving the equations of any asymptotes and the coordinates of the points where the curves meet the axes.[2] The function $\mathrm{g}$ is defined on the domain $[-1, \infty)$. The graph of $y = \mathrm{g}(x)$ has minimum point at $(2, a)$ and the equation of its asymptote is $y = b$ as shown below. Determine if $\mathrm{gf}$ exist. If it exists, find its range.[3]

Accepted Answer

(a) $a=-2$. (b) $f^{-1}(x)=\frac1{\sqrt x}-2$, $x>0$. (c) Asymptotes and intercepts shown. (d) $gf$ exists and $R_{gf}=[a,b)$, using the labels on the given $g$-graph.

2025 RVHS P1 Q10

2025 RVHS P1 Q10

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