<ol class="i-list"><li><div class="list-item-paragraph">A student defines a function $\mathrm{f}$ and its inverse ${{\mathrm{f}}^{-1}}$ by the following:</div><div class="list-item-paragraph" style="text-align: center">$\mathrm{f}:x\mapsto \sin x,$  $\qquad\,\,\,$  $x\in \mathbb{R},$  $\quad$  $-\pi \le x\le \pi $</div><div class="list-item-paragraph" style="text-align: center">${{\mathrm{f}}^{-1}}:x\mapsto {{\sin }^{-1}}x,$  $\;\;$  $x\in \mathbb{R},$  $\;\;$  $-1\le x\le 1$.</div><div class="list-item-paragraph">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.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">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.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Explain if the composite function ${{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}$ exists.<span class="inline-marks">[1]</span></div></li><li><div class="list-item-paragraph">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&nbsp;$\mathrm{gf}\left( x \right)=\frac{1}{e}$.<span class="inline-marks">[4]</span></div></li></ol>

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.

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.

Explain if the composite function ${{\mathrm{g}}^{-1}}{{\mathrm{f}}^{-1}}$ exists.

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 $\mathrm{gf}\left( x \right)=\frac{1}{e}$.

Modulus Functions

Functions

International Baccalaureate Higher Level Mathematics

IB HL Math

2016 ACJC Promo Q8

(i) The line $y = \frac{1}{2}$ cuts $\text{f}$ twice.

$\therefore$ f is not a one-one function, thus $\text{f}^{-1}$ doesn't exist.

$\therefore$ Suitable domain is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

(ii) Let $y = e^{-(x+1)^2}$, $D_{\text{g}^{-1}} = R_{\text{g}} = \left[\frac{1}{e}, 1\right]$

$\ln y = -(x+1)^2$

$-\ln y = (x+1)^2$

$x = -1 \pm \sqrt{-\ln y}$

$x = -1 + \sqrt{\ln y^{-1}}$ or $x = -1 - \sqrt{\ln y^{-1}}$

Reject $x = -1 - \sqrt{\ln y^{-1}}$ as $-1 \leq x \leq 0$

$\therefore \text{g}^{-1}: x \mapsto -1 + \sqrt{\ln\left(\frac{1}{x}\right)}, \quad x \in \mathbb{R}, \quad \frac{1}{e} \leq x \leq 1$

(iii) For $\text{g}^{-1}\text{f}^{-1}$ to exist $\Rightarrow R_{\text{f}^{-1}} \subseteq D_{\text{g}^{-1}}$

$R_{\text{f}^{-1}} = D_{\text{f}} = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$D_{\text{g}^{-1}} = \left[\frac{1}{e}, 1\right]$

$R_{\text{f}^{-1}} \not\subseteq D_{\text{g}^{-1}}$, $\therefore \text{g}^{-1}\text{f}^{-1}$ doesn't exist.

(iv) $\text{f}^{-1}\text{g}^{-1}(x) = \text{f}^{-1}\!\left[-1 + \sqrt{\ln\left(\frac{1}{x}\right)}\right] = \sin^{-1}\!\left(\sqrt{\ln\left(\frac{1}{x}\right)} - 1\right)$

$D_{\text{f}^{-1}\text{g}^{-1}} = D_{\text{g}^{-1}} = \left[\frac{1}{e}, 1\right]$

$R_{\text{g}^{-1}} = D_{\text{g}} = [-1, 0] \Rightarrow D_{\text{f}^{-1}} = [-1, 0] \Rightarrow R_{\text{f}^{-1}} = [-\pi, 0]$

Wait: $R_{\text{f}^{-1}} = \text{f}^{-1}([-1,0])$. Since $\text{f}^{-1} = \sin^{-1}$, $R_{\text{f}^{-1}\text{g}^{-1}} = \left[-\frac{\pi}{2}, 0\right]$

Given $\text{g}\text{f}(x) = \frac{1}{e}$:

$\text{g}^{-1}\text{g}\text{f}(x) = \text{g}^{-1}\!\left(\frac{1}{e}\right)$

$\text{f}(x) = \text{g}^{-1}\!\left(\frac{1}{e}\right)$

$\text{f}^{-1}\text{f}(x) = \text{f}^{-1}\text{g}^{-1}\!\left(\frac{1}{e}\right)$

$x = \text{f}^{-1}\text{g}^{-1}\!\left(\frac{1}{e}\right) = \sin^{-1}\!\left(\sqrt{\ln\frac{1}{\frac{1}{e}}} - 1\right) = \sin^{-1}(1 - 1) = \sin^{-1}(0) = 0$

$\therefore x = 0$

2016 ACJC Promo Q8

2016 ACJC Promo Q8

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