Given the functions $\mathrm{f}:x\mapsto 3{{x}^{3}}-4{{x}^{2}}-1$, $x\ge 1$, $\mathrm{g}:x\mapsto {{

Question

Given the functions $\mathrm{f}:x\mapsto 3{{x}^{3}}-4{{x}^{2}}-1$, $x\ge 1$, $\mathrm{g}:x\mapsto {{\mathrm{e}}^{2-0.25{{x}^{2}}}}$, $x\ge -3$. Explain why the functions ${{\mathrm{f}}^{-1}}$ and $\mathrm{gf}$ exist, [2] Find the exact value of ${{\mathrm{f}}^{-1}}\left( -1 \right)$, [2] Find the range of $\mathrm{gf}$, giving your answer in exact form. [2] The functions $\mathrm{f}$ and $\mathrm{h}$ are defined by $\mathrm{f}:x\mapsto {{x}^{2}}-{{\lambda }^{2}}$, $x\le -\lambda $, $\mathrm{h}:x\mapsto \sqrt{x+{{\lambda }^{2}}}$, $x\ge 0$. where $\lambda >0$. Define the following functions ${{\mathrm{f}}^{-1}}$. $\mathrm{hf}$. [5]

Accepted Answer

$R_{gf} = (0, e^2]$; $f^{-1}: x \mapsto -\sqrt{x + \lambda^2}$; $hf: x \mapsto x$

2010 ACJC CT2

2010 ACJC CT2

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