Values of x that make the denominator equal to zero

Oct 13, 2025 by pinkandtrash.com 45 views

Introduction

In this article, we will explore the functions $f$ and $g$ , which are defined as follows:

${ \begin{align*} f(x) &= \frac{x-7}{x^2+14x+49} \\ g(x) &= \frac{x-3}{x^2-x-6} \end{align*} \}$

Our goal is to find the domain of each function, which is the set of all possible input values for which the function is defined.

Function f

To find the domain of $f$ , we need to consider the values of $x$ that make the denominator of the function equal to zero.

Denominator of f

The denominator of $f$ is given by:

${ x^2 + 14x + 49 }$

We can factor the denominator as:

${ (x + 7)^2 }$

Values of x that make the denominator equal to zero

To find the values of $x$ that make the denominator equal to zero, we set the denominator equal to zero and solve for $x$ :

${ (x + 7)^2 = 0 }$

Solving for $x$ , we get:

${ x + 7 = 0 }$

${ x = -7 }$

Therefore, the value of $x$ that makes the denominator equal to zero is $x = -7$ .

Domain of f

Since the denominator of $f$ is equal to zero when $x = -7$ , we need to exclude this value from the domain of $f$ .

The domain of $f$ is therefore:

${ (-\infty, -7) \cup (-7, \infty) }$

Function g

To find the domain of $g$ , we need to consider the values of $x$ that make the denominator of the function equal to zero.

Denominator of g

The denominator of $g$ is given by:

${ x^2 - x - 6 }$

We can factor the denominator as:

${ (x - 3)(x + 2) }$

Values of x that make the denominator equal to zero

To find the values of $x$ that make the denominator equal to zero, we set the denominator equal to zero and solve for $x$ :

${ (x - 3)(x + 2) = 0 }$

Solving for $x$ , we get:

${ x - 3 = 0 }$

${ x = 3 }$

${ x + 2 = 0 }$

${ x = -2 }$

Therefore, the values of $x$ that make the denominator equal to zero are $x = 3$ and $x = -2$ .

Domain of g

Since the denominator of $g$ is equal to zero when $x = 3$ and $x = -2$ , we need to exclude these values from the domain of $g$ .

The domain of $g$ is therefore:

${ (-\infty, -2) \cup (-2, 3) \cup (3, \infty) }$

Conclusion

In this article, we have found the domain of the functions $f$ and $g$ . The domain of $f$ is $(-\infty, -7) \cup (-7, \infty)$ , and the domain of $g$ is $(-\infty, -2) \cup (-2, 3) \cup (3, \infty)$ .

References

[1] "Algebra and Trigonometry" by Michael Sullivan
[2] "Calculus" by James Stewart

Discussion

Introduction

In our previous article, we explored the functions $f$ and $g$ , which are defined as follows:

${ \begin{align*} f(x) &= \frac{x-7}{x^2+14x+49} \\ g(x) &= \frac{x-3}{x^2-x-6} \end{align*} \}$

We found the domain of each function, which is the set of all possible input values for which the function is defined.

In this article, we will answer some common questions about the functions $f$ and $g$ .

Q&A

Q: What is the domain of f?

A: The domain of $f$ is $(-\infty, -7) \cup (-7, \infty)$ .

Q: What is the domain of g?

A: The domain of $g$ is $(-\infty, -2) \cup (-2, 3) \cup (3, \infty)$ .

Q: Why do we need to exclude certain values from the domain of a function?

A: We need to exclude certain values from the domain of a function because these values make the denominator of the function equal to zero. When the denominator is equal to zero, the function is undefined.

Q: How do we find the values that make the denominator equal to zero?

A: To find the values that make the denominator equal to zero, we set the denominator equal to zero and solve for $x$ .

Q: What happens if we try to evaluate a function at a value that is not in its domain?

A: If we try to evaluate a function at a value that is not in its domain, we will get an undefined result.

Q: Can we always find the domain of a function?

A: No, we cannot always find the domain of a function. For example, if the denominator of a function is a polynomial of degree greater than 1, it may be difficult or impossible to find the values that make the denominator equal to zero.

Q: How does the domain of a function affect the graph of the function?

A: The domain of a function affects the graph of the function by determining the set of points that are included in the graph. If a value is not in the domain of a function, it will not be included in the graph.

Q: Can we always graph a function?

A: No, we cannot always graph a function. If a function has a complex or undefined domain, it may be difficult or impossible to graph.

Conclusion

In this article, we have answered some common questions about the functions $f$ and $g$ . We have discussed the domain of each function, how to find the values that make the denominator equal to zero, and how the domain of a function affects the graph of the function.

References

[1] "Algebra and Trigonometry" by Michael Sullivan
[2] "Calculus" by James Stewart

Discussion

What are some other questions you have about the functions $f$ and $g$ ? How do you think the domain of a function affects the graph of the function?