Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.1 Functions - Exercises Set 0.1: 10

Answer

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

Work Step by Step

(a) 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 $f(x)= \sqrt{3-x}$ is$$D_f=\mathbb R - \{ x \mid 3-x<0 \}= \mathbb R - \{ x \mid x>3 \} \\ = ( - \infty , 3].$$To find the range of $f(x)=\sqrt{3-x}$, one should note that the range of $f_1(x)= \sqrt{x}$ is $[0, \infty )$ and the range of $f_2(x)=3-x$ is $\mathbb R$. So the range of $f(x)=\sqrt{3-x}$ is $$R_f=[0, \infty ).$$ (b) 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 $F(x)=\sqrt{4-x^2}$ is$$D_F= \mathbb R - \{ x \mid 4-x^2<0 \}= \mathbb R - \{ x \mid x2 \} \\= [-2, 2].$$To find the range of $F(x)=\sqrt{4-x^2}$, one should note that$$x^2 \ge 0 \quad \Rightarrow \quad 4-x^2 \le 4 \quad \Rightarrow \quad 0 \le \sqrt{4-x^2} \le 2.$$So the range of $F$ is$$R_F=[0,2].$$ (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 $g(x)=3+ \sqrt{x}$ is$$D_g= \mathbb R - \{ x \mid x<0 \}= [0, \infty ).$$To find the range of $g(x)=3+ \sqrt{x}$, one should note that the range of $f_1(x)= \sqrt{x}$ is $[0, \infty )$. So the range of $g(x)=3+ \sqrt{x}$ is$$R_g=[3, \infty ).$$ (d) Domain of real polynomial function is the set of real numbers, $\mathbb R$. So the domain of $G(x)= x^3+2$ is$$D_G= \mathbb R.$$To find the range of $G$, one should note that the value of $x^3$ can be any real number. So the range of $G(x)=x^3+2$ is$$R_G= \mathbb R.$$ (e) The domain of sine function, $\sin x$, is the set of real numbers, $\mathbb R$. So the domain of $h(x)= 3 \sin x$ is$$D_h= \mathbb R.$$To find the range of $h$, one should note that $-1 \le \sin x \le 1$. So the range of $h(x)= 3 \sin x$ is$$R_h=[-3, 3].$$ (f) The domain of sine function, $\sin x$, is the set of real numbers, $\mathbb R$, domain of a real root function $\sqrt{f}$ is the set of real numbers, $\mathbb R$, minus those points which make $f$ negative, and domain of a real rational function is the set of real numbers, $\mathbb R$, minus those points vanishing its denominator. So the domain of $H(x)=( \sin \sqrt{x})^{-2}= \left ( \frac{1}{ \sin \sqrt{x}} \right )^2$ is$$D_H= \mathbb R - \{ x \mid x<0 \} - \{ x \mid \sqrt{x}= k \pi , k \in \mathbb Z \} \\ = (0, \infty ) - \{ x \mid x = k^2 \pi ^2 , k \in \mathbb Z \}.$$To find the range of $H$, one should note that $-1 \le \sin x \le 1$. So$$\left ( \frac{1}{ \sin \sqrt{x}} \right ) \le -1 \, \text{ or } \, \left ( \frac{1}{ \sin \sqrt{x}} \right ) \ge 1$$ $$\Rightarrow \quad \left ( \frac{1}{ \sin \sqrt{x}} \right )^2 \ge 1.$$Thus, the range of $H(x)=( \sin \sqrt{x})^{-2}$ is$$R_H=[1, \infty ).$$
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