Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.3 Functions - 2.3 Exercises: 46

Answer

The given relation defines $y$ as a function of $x$. domain: $(-\infty, 3.5]$ range: $[0, +\infty)$

Work Step by Step

The given equation above will give only one value of $y$ for every value of $x$ within its domain. 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 $\sqrt{7-2x}$, the value of the radicand (which in this case is $7-2x$) cannot be negative. Thus, $7-2x \ge 0 \\-2x \ge -7 \\\frac{-2x}{-2} \le \frac{-7}{-2} \\x \le 3.5$ This means that the value of $x$ can be any number less than or equal to $3.5$. Thus, the domain is $(-\infty, 3.5]$. Note that $\sqrt{7-2x}$ represents the principal square root of $7-2x$. 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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