## College Algebra (10th Edition)

Published by Pearson

# Chapter 3 - Section 3.1 - Functions - 3.1 Assess Your Understanding - Page 211: 69

#### Answer

$\begin{array}{llllll} & & domain & & & \\ a. & 2x^{2}+x-1, & x\in \mathbb{R} & \text{ }& e. & 20\\ b. & -2x^{2}+x-1, & x\in \mathbb{R} & & f. & -29\\ c. & 2x^{3}-2x^{2}, & x\in \mathbb{R} & & g. & 8\\ d. & \dfrac{x-1}{2x^{2}}, & x\neq 0 & & h. & 0 \end{array}$

#### Work Step by Step

$f(x)=x-1 g(x)=2x^{2}$ $\mathrm{a}.\ (f+g)(x)=f(x)+g(x)=$ $=x-1+2x^{2}$ $=2x^{2}+x-1$ Domain: $\{x|x \in \mathbb{R}\}$. $\mathrm{b}.\ (f-g)(x)=f(x)-g(x)$ $=(x-1)-(2x^{2})$ $=x-1-2x^{2}$ $=-2x^{2}+x-1$ Domain: $\{x|x \in \mathbb{R}\}$. $\mathrm{c}.\ (f\cdot g)(x)=f(x)\cdot g(x)$ $=(x-1)(2x^{2})$ $=2x^{3}-2x^{2}$ Domain: $\{x|x \in \mathbb{R}\}$. $\mathrm{d}.\ (\displaystyle \frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{x-1}{2x^{2}}$ Domain: $\{x|x\neq 0\}$. $\mathrm{e}.\ (f+g)(3)=$... substitute x=3 in the result of a. $=2(3)^{2}+3-1$ $=2(9)+3-1$ $=18+3-1$ $=20$ $\mathrm{f}.\ (f-g)(4)=$... substitute x=4 in the result of b. $=-2(4)^{2}+4-1$ $=-2(16)+4-1$ $=-32+4-1=-29$ $\mathrm{g}.\ (f\cdot g)(2)=$ ... substitute x=2 in the result of c. $=2(2)^{3}-2(2)^{2}$ $=2(8)-2(4)$ $=16-8$ $=8$ $\mathrm{h}.\ (\displaystyle \frac{f}{g})(1)=$... substitute x=1 in the result of d. $=\displaystyle \frac{1-1}{2(1)^{2}}$ $=\displaystyle \frac{0}{2(1)}$ $=0$

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