/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Find the domain of \(f\) $$ ... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 38

Find the domain of \(f\) $$ f(x)=\frac{3 x}{x^{2}+1} $$

Short Answer

Expert verified
The domain of \( f(x) \) is all real numbers.

Step by step solution

01

Identify the Function

The function given is \[ f(x) = \frac{3x}{x^2 + 1} \]Identify any restrictions on the domain based on the denominator.
02

Analyzing the Denominator

To ensure the function is defined, the denominator must not be zero. So, analyze the denominator:\[ x^2 + 1 \]Solve for zero to find potential restrictions: \[ x^2 + 1 = 0 \]
03

Solve for Zero

Solve the equation \[ x^2 + 1 = 0 \]This simplifies to \[ x^2 = -1 \]which has no real solutions since the square of any real number is always non-negative.
04

Conclusion about the Domain

Since the denominator \[ x^2 + 1 \]never equals zero, the function \[ f(x) \]is defined for all real numbers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Analyzing the Denominator
To find the domain of a function, we need to ensure that the function is defined for all inputs. In rational functions like \(f(x) = \frac{3x}{x^2 + 1}\), the denominator should not be zero. When we analyze the denominator \(x^2 + 1\), we set it equal to zero to check for any values that might make the function undefined.

Solving \(x^2 + 1 = 0\) leads to \(x^2 = -1\), which is impossible for real numbers because no real number squared gives a negative result. Thus, the denominator never zeroes out, meaning there are no restrictions from the denominator.

Hence, the function is defined for all real numbers.
Real Numbers
A key part of understanding the domain of a function is knowing what values are included in the set of real numbers. Real numbers include all the numbers on the number line:
  • Positive numbers (e.g., 1, 2.5, 3/4)
  • Negative numbers (e.g., -1, -2.5, -3/4)
  • Zero


Real numbers do not include imaginary or complex numbers, which are used when dealing with the square roots of negative numbers. Since the denominator of our function never reaches zero over the set of real numbers, our function is defined over this entire set.

Hence, the domain of \( f(x) = \frac{3x}{x^2 + 1} \) is all real numbers.
Function Definition
To clearly answer the question about finding the domain of a function, let鈥檚 define what a domain is:

The domain of a function is the complete set of possible values of the independent variable. In simpler terms, it represents all the x-values that you can plug into the function without causing any undefined behavior like division by zero.

For the given function \( f(x) = \frac{3x}{x^2 + 1} \), defining the domain means checking which x-values make the function work properly. Here:
  • The denominator \( x^2 + 1 \) never equals zero because \( x^2 \) (square of any real number) is always non-negative, and adding 1 makes it positive.
  • Since the function doesn鈥檛 have a denominator that hits zero, it's defined for all x in the set of real numbers.


Therefore, the domain of \( f(x) \) is \( (-鈭�, 鈭�) \), meaning every real number can be an input for this function.

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Most popular questions from this chapter

To prepare for Section 9.2, review translating sentences to equations (Section 1.1). Translate each sentence to an equation.\([1.1]\) The sum of three numbers is \(100 .\)

For each of the following pairs of total-cost and total revenue functions, find (a) the total-profit function and (b) the break-even point. $$ \begin{array}{l} {C(x)=30 x+49,500} \\ {R(x)=85 x} \end{array} $$

Find the domain of \(f\) $$ f(x)=\frac{x-5}{2 x+1} $$

Alejandro receives \(\$ 306\) per year in simple interest from three investments totaling \(\$ 3200\). Part is invested at \(8 \%\), part at \(9 \%,\) and part at \(10 \% .\) There is \(\$ 1900\) more invested at \(10 \%\) than at \(9 \% .\) Find the amount invested at each rate.

For each of the following pairs of total-cost and total revenue functions, find (a) the total-profit function and (b) the break-even point. $$ \begin{array}{l} {C(x)=20 x+120,000} \\ {R(x)=50 x} \end{array} $$
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