Finding the Domain of Rational Functions
Welcome to Professor Baker's Math Class! Today, we're diving into the domain of rational functions. Understanding the domain is crucial because it tells us all the possible input values (x-values) that a function can accept without resulting in undefined operations, like division by zero.
What is a Rational Function?
A rational function is a function that can be written in the form:
$$f(x) = \frac{N(x)}{D(x)}$$where $N(x)$ and $D(x)$ are polynomials, and crucially, $D(x) \neq 0$. The denominator cannot equal zero, as division by zero is undefined. A rational number can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. The number 1.18325 can be expressed as $\frac{118325}{100000}$.
Domain Restrictions
When determining the domain of a rational function, we need to consider the following restrictions:
- Division by Zero: The denominator, $D(x)$, cannot be zero. We must exclude any $x$-values that make $D(x) = 0$.
- Even Roots: If the function involves even roots (square root, fourth root, etc.), the expression inside the root must be greater than or equal to zero. However, this restriction is not directly related to rational functions, but is still important to keep in mind.
- Real-Life Restrictions: In some real-world applications, there might be additional restrictions based on the context of the problem.
Finding the Domain: Examples
Let's look at some examples to illustrate how to find the domain of rational functions:
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$f(x) = \frac{1}{x}$
Here, the denominator is $x$. We must have $x \neq 0$. Therefore, the domain is all real numbers except 0, which can be written as $(-\infty, 0) \cup (0, \infty)$.
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$f(x) = \frac{3x - 5}{x^2 - 4}$
The denominator is $x^2 - 4$. We need to find the values of $x$ for which $x^2 - 4 \neq 0$. We solve $x^2 - 4 = 0$, which gives us $(x - 2)(x + 2) = 0$. Thus, $x = 2$ and $x = -2$ must be excluded. The domain is all real numbers except 2 and -2, or $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$.
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$f(x) = \frac{1}{x^2 + 4}$
The denominator is $x^2 + 4$. We need to find the values of $x$ for which $x^2 + 4 \neq 0$. In this case, $x^2 + 4$ is always greater than 0 for any real number $x$, since $x^2$ is always non-negative. Therefore, there are no restrictions, and the domain is all real numbers, $(-\infty, \infty)$.
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$f(x) = \frac{3x}{x^2 - 6x + 9}$
The denominator is $x^2 - 6x + 9$. We need to find the values of $x$ for which $x^2 - 6x + 9 \neq 0$. We solve $x^2 - 6x + 9 = 0$, which factors to $(x - 3)(x - 3) = 0$. Thus, $x = 3$ must be excluded. The domain is all real numbers except 3, or $(-\infty, 3) \cup (3, \infty)$.
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$f(x) = \frac{7x^2 - 15x + 1592}{|x| - 2}$
The denominator is $|x| - 2$. We need to find the values of $x$ for which $|x| - 2 \neq 0$. We solve $|x| - 2 = 0$, which gives us $|x| = 2$. Thus, $x = 2$ and $x = -2$ must be excluded. The domain is all real numbers except 2 and -2, or $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$.
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$f(x) = \frac{10}{|x+2| - 5}$
The denominator is $|x+2| - 5$. We need to find the values of $x$ for which $|x+2| - 5 \neq 0$. We solve $|x+2| - 5 = 0$, which gives us $|x+2| = 5$. Thus, $x+2 = 5$ or $x+2 = -5$, which means $x = 3$ and $x = -7$ must be excluded. The domain is all real numbers except 3 and -7, or $(-\infty, -7) \cup (-7, 3) \cup (3, \infty)$.
Homework
Practice makes perfect! Complete the following problems from your textbook to solidify your understanding:
- Page 149, problems #13-22. Remember to find the domain only!
Keep up the great work, and see you in the next class!