## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - 2.3 Functions - 2.3 Exercises - Page 216: 45

#### Answer

The given relation defines $y$ as a function of $x$. domain: $[-\frac{1}{4}, \infty)$ range: $[0, +\infty)$

#### Work Step by Step

The given equation above will give only one value of $y$ for every value of $x$. This means that each $x$ is paired with only one value of $y$. Thus, the given relation defines $y$ as a function of $x$. Note that in $\sqrt{4x+1}$, the value of the radicand (which in this case is $4x+2$) cannot be negative. Thus, $4x+1 \ge 0 \\4x \ge -1 \\x \ge \frac{-1}{4}$ This means that the value of $x$ can be any number greater than or equal to $-\frac{1}{4}$. Thus, the domain is $[-\frac{1}{4}, \infty)$. Note that $\sqrt{4x+1}$ represents the principal square root of $4x+1$. The principal square root of a number is either zero or positive. Thus, the value of $y$ can be any real number greater than or equal to zero. In interval notation, the range is $[0, +\infty)$.

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