Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.7 - Combining Functions - 2.7 Exercises - Page 216: 8

Answer

$f+g=x+\sqrt{x}$ $D=\{x|x\geq0\}$ or $D=[0,\infty)$ $f-g=x-\sqrt{x}$ $D=\{x|x\geq0\}$ or $D=[0,\infty)$ $fg=x\sqrt{x}$ $D=\{x|x\geq0\}$ or $D=[0,\infty)$ $f/g=\dfrac{x}{\sqrt{x}}$ $D=\{x|x\gt0\}$ or $D=(0,\infty)$

Work Step by Step

$f(x)=x,$ $g(x)=\sqrt{x}$ $f+g$ $f(x)+g(x)=x+\sqrt{x}$ The domain of this function doesn't include the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain of this function only has the numbers greater than or equal to $0$. The domain can be expressed as follows: $D=\{x|x\geq0\}$ or $D=[0,\infty)$ $f-g$ $f(x)-g(x)=x-\sqrt{x}$ The domain of this function doesn't include the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain of this function only has the numbers greater than or equal to $0$. The domain can be expressed as follows: $D=\{x|x\geq0\}$ or $D=[0,\infty)$ $fg$ $f(x)\cdot g(x)=x\sqrt{x}$ The domain of this function doesn't include the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain of this function only has the numbers greater than or equal to $0$. The domain can be expressed as follows: $D=\{x|x\geq0\}$ or $D=[0,\infty)$ $f/g$ $\dfrac{f(x)}{g(x)}=\dfrac{x}{\sqrt{x}}$ This function is undefined for the values of $x$ that make the denominator equal to $0$ and for the values of $x$ that make the expression inside the square root a negative number. Therefore, the domain only includes the numbers greater than $0$. The domain can be expressed as follows: $D=\{x|x\gt0\}$ or $D=(0,\infty)$
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