The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{\left( x+3 \right)}^{2}}-1$ for&n

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto {{\left( x+3 \right)}^{2}}-1$ for $x\in \mathbb{R}$, $x\ge -3$. Find ${{\mathrm{f}}^{-1}}\left( x \right)$, stating the domain of ${{\mathrm{f}}^{-1}}$.[3] The function $\mathrm{g}$ is defined by $\mathrm{g}\left( x \right)=\left\{ \begin{matrix} x-1\,\,\,\,\,\,\mathrm{for}\,\,\,\,0<x<1\, \ {{x}^{2}}-1\,\,\,\,\mathrm{for}\,\,1\le x\le 2 \ \end{matrix} \right.$ and $\mathrm{g}\left( x \right)=\mathrm{g}\left( x+2 \right)$ for all values of $x$. Sketch the graph of $\mathrm{g}$ for $-1\le x<3$. Hence state the range of $\mathrm{g}$.[3] The domain of $\mathrm{g}$ is now restricted to $-1\le x<1$. Explain why the composite function ${{\mathrm{f}}^{-1}}\mathrm{g}$ exists.[1] Find ${{\mathrm{f}}^{-1}}\mathrm{g}\left( x \right)$ in the form ${{\mathrm{f}}^{-1}}\mathrm{g}\left( x \right)=\left\{ \begin{matrix} \mathrm{p}\left( x \right)\,\,\,\,\mathrm{for}\,\,-1<x\le 0 \ \mathrm{q}\left( x \right)\,\,\,\,\mathrm{for}\,\,\,\,0<x<1 \ \end{matrix} \right.$ where $\mathrm{p}\left( x \right)$ and $\mathrm{q}\left( x \right)$ are expressions in terms of $x$ to be determined.[3]

Accepted Answer

(a) $f^{-1}(x)=-3+\sqrt{x+1}$, $D_{f^{-1}}=[-1,\infty)$ (b) $R_g=(-1,3]$ (c)(i) $R_g\subseteq D_{f^{-1}}$, so $f^{-1}g$ exists. (c)(ii) $p(x)=x-1$, $q(x)=-3+\sqrt{x}$

2022 EJC P1 Q4

2022 EJC P1 Q4

Video Solution:

Solution:

Solution locked