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: 10

Answer

$f+g=-1$ $D:\{x|x\in R\}$ or $D:(-\infty,\infty)$ $f-g=-2x^{2}+7$ $D:\{x|x\in R\}$ or $D:(-\infty,\infty)$ $fg=-x^{4}+7x^{2}-12$ $D:\{x|x\in R\}$ or $D:(-\infty,\infty)$ $f/g=\dfrac{3-x^{2}}{x^{2}-4}$ $D:\{x|x\ne2,-2\}$ or $D:(-\infty,-2)\cup(-2,2)\cup(2,\infty)$

Work Step by Step

$f(x)=3-x^{2},$ $g(x)=x^{2}-4$ $f+g$ $f(x)+g(x)=3-x^{2}+x^{2}-4=-1$ The domain of this function includes all real numbers. It can be expressed as follows: $D=\{x|x\in R\}$ or $D=(-\infty,\infty)$ $f-g$ $f(x)-g(x)=3-x^{2}-(x^{2}-4)=3-x^{2}-x^{2}+4=-2x^{2}+7$ The domain of this function includes all real numbers. It can be expressed as follows: $D=\{x|x\in R\}$ or $D=(-\infty,\infty)$ $fg$ $f(x)\cdot g(x)=(3-x^{2})(x^{2}-4)=3x^{2}-12-x^{4}+4x^{2}=...$ $...=-x^{4}+7x^{2}-12$ The domain of this function includes all real numbers. It can be expressed as follows: $D=\{x|x\in R\}$ or $D=(-\infty,\infty)$ $f/g$ $\dfrac{f(x)}{g(x)}=\dfrac{3-x^{2}}{x^{2}-4}$ This function is undefined for the values of $x$ that make the denominator equal to $0$. Let's find these values solving the following equation: $x^{2}-4=0$ $x^{2}=4$ $\sqrt{x^{2}}=\sqrt{4}$ $x=\pm2$ Now we know the values of $x$ for which the function is undefined. The domain can be expressed as follows: $D:\{x|x\ne2,-2\}$ or $D:(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
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