Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.1 Rational Functions and Variation - 7.1 Exercises - Page 566: 45

Answer

The domain of this function is all real numbers except $-3$ and $-2$.

Work Step by Step

To find the domain of this function, we need to find which values are excluded for $r$. In a rational function, the denominator cannot equal $0$ because the function would be undefined. Therefore, we need to set the denominator equal to $0$ and solve for $r$: $r^2 + 5r + 6 = 0$ We can factor this equation to solve for $r$. We see that we have a quadratic equation, which is given by the formula: $ax^2 + bx + c$, where $a$, $b$, and $c$ are all real numbers. To factor this equation, we want to find which factors when multiplied will give us the product of the $a$ and $c$ terms, which is $6$, but when added together will give us the $b$ term, which is $5$. This means that both factors should be positive. Let's look at possible factors: $3$ and $2$ $6$ and $-1$ It looks like the first combination will work. Let's put the factors together: $(r + 3)(r + 2) = 0$ According to the zero product property, if the product of two factors equals $0$, then either factor can be $0$; therefore, we can set each of these factors equal to $0$ and solve: $r + 3 = 0$ or $r + 2 = 0$ Add or subtract to solve: $r = -3$ or $r = -2$ These are the numbers $r$ cannot be. Therefore, the domain of this function is all real numbers except $-3$ and $-2$.
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