## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 0 - Before Calculus - 0.1 Functions - Exercises Set 0.1 - Page 13: 9

#### Answer

(a)$$D_f=\mathbb R - \{ 3 \}$$ $$R_f=\mathbb R - \{ 0 \}$$ (b)$$D_F=\mathbb R - \{ 0 \}$$ $$R_F=\{-1, 1 \}$$ (c)$$D_g=( - \infty , - \sqrt{3} ] \cup [ \sqrt{3} , \infty )$$ $$R_g= [0, \infty )$$ (d)$$D_G= \mathbb R$$ $$R_G=[2, \infty )$$ (e)$$D_h=\mathbb R - \{ x \mid x= \frac{ \pi}{2}+ 2k \pi, k \in \mathbb Z \}$$ $$R_h= \left[ \frac{1}{2} , \infty \right)$$ (f)$$D_H=[-2, 2) \cup (2, \infty )$$ $$R_H= [0, 2) \cup (2, \infty )$$

#### Work Step by Step

(a) Domain of a real rational function is the set of real numbers, $\mathbb R$, minus those points vanishing its denominator. So the domain of $f(x)=\frac{1}{x-3}$ is$$D_f=\mathbb R - \{ 3 \}.$$ To find the range of $f$, suppose $y \in \mathbb R$ is an arbitrary point in the range of $f$, so there exists a point $x \in D_f$ such that $y=f(x)$. Then we have$$y= \frac{1}{x-3} \quad \Rightarrow \quad yx-3y=1 \quad \Rightarrow \quad x=\frac{1+3y}{y}.$$ The last fraction is defined for all $y \in \mathbb R$ except $y=0$. Thus, the range of $f$ is $$R_f=\mathbb R - \{ 0 \}.$$ (b) Domain of a real rational function is the set of real numbers, $\mathbb R$, minus those points vanishing its denominator. So the domain of $F(x)=\frac{x}{|x|}$ is $$D_F= \mathbb R - \{ 0 \}.$$To find the range of $F$, suppose $y \in \mathbb R$ is an arbitrary point in the range of $F$, so there exists a point $x \in D_F$ such that $y=F(x)$. Then we have$$y=\frac{x}{|x|} \quad \Rightarrow \quad y|x|=x$$ $$= \begin{cases} yx=x & \text{ if } x>0 \\ y(-x)=x & \text{ if } x<0 \end{cases} = \begin{cases} y=1 & \text { if } x>0 \\ y=-1 & \text{ if } x<0 \end{cases}$$Thus, the range of $F$ is $$R_F=\{-1, 1 \}.$$ (c) Domain of a real root function $\sqrt{f}$ is the set of real numbers, $\mathbb R$, minus those points which make $f$ negative. So the domain of $g(x)= \sqrt{x^2-3}$ is$$D_g= \mathbb R - \{ x \mid x^2-3<0 \}= \mathbb R - \{ x \mid - \sqrt{3} < x < \sqrt{3} \}$$ $$=( - \infty , - \sqrt{3} ] \cup [ \sqrt{3} , \infty ).$$To find the range of $g(x)= \sqrt{x^2-3}$, one should note that the range of $f_1(x)= \sqrt{x}$ is $[0, \infty )$ and the range of $f_2(x)=x^2-3$ is $[-3, \infty )$. So the range of $g(x)= \sqrt{ x^2 -3}$ is$$R_g= [0, \infty ).$$ (d) Domain of a real root function $\sqrt{f}$ is the set of real numbers, $\mathbb R$, minus those points which make $f$ negative. So the domain of $G(x)= \sqrt{x^2-2x+5}$ is$$D_G= \mathbb R - \{ x \mid x^2-2x+5<0 \} = \mathbb R - \varnothing = \mathbb R$$ (Note that $x^2-2x+5=(x-1)^2+4$ never becomes negative). To find the range of $g(x)= \sqrt{x^2-2x+5}$, one should note that the range of $f_1(x)= \sqrt{x}$ is $[0, \infty )$ and the range of $f_2(x)=x^2-2x+5=(x-1)^2+4$ is $[4, \infty )$. So the range of $G(x)= \sqrt{x^2-2x+5}$ is$$R_G=[2, \infty ).$$ (e) Domain of a real rational function is the set of real numbers, $\mathbb R$, minus those points vanishing its denominator, and the domain of the sine function is $\mathbb R$. So the domain of $h(x)= \frac{1}{1- \sin x}$ is$$D_h=\mathbb R - \{ x \mid 1 - \sin x =0 \} \\ = \mathbb R - \{ x \mid x= \frac{ \pi}{2}+ 2k \pi, k \in \mathbb Z \}.$$To find the range of $h$, one should note that $-1 \le \sin x \le 1$, so we have$$0 \le 1- \sin x \le 2 \quad \Rightarrow \quad \frac{1}{1 - \sin x} \ge \frac{1}{2}.$$Thus, the range of $h$ is$$R_h= [ \frac{1}{2} , \infty ).$$ (f) Domain of a real rational function is the set of real numbers, $\mathbb R$, minus those points vanishing its denominator, and domain of a real root function $\sqrt{f}$ is the set of real numbers, $\mathbb R$, minus those points which make $f$ negative. So the domain of $H(x)= \sqrt{ \frac{x^2 -4}{x-2}}$ is$$D_H=[-2, \infty )- \{ 2 \} = [-2, 2) \cup (2, \infty )$$(Note that for $x \in \mathbb R - \{ 2 \}$ the fraction under the root can be written as $\frac{(x+2)(x-2)}{x-2}=x+2$). To find the range of $H(x)= \sqrt{ \frac{x^2 -4}{x-2}}= \sqrt{x+2}$ (for $x \in D_H$), one should note that the range of $f_1(x)= \sqrt{x}$ is $[0, \infty )$ and the range of $f_2(x)=x+2$ is $\mathbb R$. Thus, the range of $H$ is$$R_H= [0, \infty ) - \{ 2 \}= [0, 2) \cup (2, \infty )$$(Note that one must exclude the value of $\sqrt{x+2}$ for $x=2$ since $x=2$ is not in the domain of $H$).

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