A function $\mathrm{g}$ is defined by \({\mathrm{g}(x) = \begin{cases} \frac{2}{1+x^2} & \text{if}

Question

A function $\mathrm{g}$ is defined by ${\mathrm{g}(x) = \begin{cases} \frac{2}{1+x^2} & \text{if}\hspace{0.5em} x \leq 0, \ 0 & \text{if}\hspace{0.5em} x > 0. \end{cases}}$ Give a reason why $\mathrm{g}$ does not have an inverse.[1] The function ${{\mathrm{g}}^{-1}}$ exists if the domain of $\mathrm{g}$ is restricted to $x\le k$. State the greatest possible value of $k$.[1] In the rest of the question, the domain of $\mathrm{g}$ is $x\le k$, where $k$ takes the value determined in part (b). Find ${{\mathrm{g}}^{-1}}\left( x \right)$ and state the domain of ${{\mathrm{g}}^{-1}}$.[3] Sketch, on the axes given below, the graphs of $y=\mathrm{g}\left( x \right)$ and $y={{\mathrm{g}}^{-1}}\left( x \right)$. Label the two graphs clearly. Write down the equation of the line in which the graph of $y=\mathrm{g}\left( x \right)$ must be reflected in order to obtain the graph of $y={{\mathrm{g}}^{-1}}\left( x \right)$.[3]

Accepted Answer

(a) $\mathrm g$ is not one-to-one, e.g. $\mathrm g(1)=\mathrm g(2)=0$. (b) $k=0$. (c) $\mathrm g^{-1}(x)=-\sqrt{\frac2x-1}$, with domain $0<x\le2$. (d) Reflection line: $y=x$.

2023 EJC Promo Q7

2023 EJC Promo Q7

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