A student defines a function $\mathrm{f}$ and its inverse ${{\mathrm{f}}^{-1}}$ by the following

Question

A student defines a function $\mathrm{f}$ and its inverse ${{\mathrm{f}}^{-1}}$ by the following: $\mathrm{f}:x\mapsto \sin x,$ $\qquad\,\,\,$ $x\in \mathbb{R},$ $\quad$ $-\pi \le x\le \pi $ ${{\mathrm{f}}^{-1}}:x\mapsto {{\sin }^{-1}}x,$ $\;\;$ $x\in \mathbb{R},$ $\;\;$ $-1\le x\le 1$. With the aid of a sketch, explain what is wrong with the definition of $\mathrm{f}$, and suggest a suitable restriction to its domain such that the definition of ${{\mathrm{f}}^{-1}}$ is correct.[2] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto {{e}^{-{{\left( x+1 \right)}^{2}}}}$, $x\in \mathbb{R}$, $-1\le x\le 0$. Find the inverse function ${{\mathrm{g}}^{-1}}$ in similar form.[3] Explain if the composite function ${{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}$ exists.[1] Write down ${{\mathrm{f}}^{-1}}{{\mathrm{g}}^{-1}}\left( x \right)$and its domain, and find its range. Hence, or otherwise, find the value of $x$ such that $\mathrm{gf}\left( x \right)=\frac{1}{e}$.[4]

Accepted Answer

(i) $\displaystyle \mathrm{D}_{\mathrm{f}}=\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (ii) $\displaystyle \mathrm{g}^{-1}:x\mapsto -1+\sqrt{\ln\left(\frac{1}{x}\right)},\ \frac{1}{e}\le x\le1$ (iii) Does not exist (iv) $\displaystyle \mathrm{f}^{-1}\mathrm{g}^{-1}(x)=\sin^{-1}\left(\sqrt{\ln\left(\frac{1}{x}\right)}-1\right)$, domain $\displaystyle \left[\frac{1}{e},1\right]$, range $\displaystyle \left[-\frac{\pi}{2},0\right]$, $x=0$

2016 ACJC Promo Q8

2016 ACJC Promo Q8

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