Domain Of A Rational Function F(x)=(5x-4)/(12x^2+17x-5) In Interval Notation

#h1 Understanding the Domain of Functions

The domain of a function is a fundamental concept in mathematics, defining the set of all possible input values (often denoted as x) for which the function produces a valid output. In simpler terms, it's the collection of all x-values that you can plug into a function without causing any mathematical errors, such as division by zero or taking the square root of a negative number. Identifying the domain is crucial because it helps us understand the function's behavior and limitations. For a given function, determining its domain often involves considering several factors, including the presence of fractions, radicals, and logarithms. Each of these elements can impose specific restrictions on the input values. For instance, if a function contains a fraction, the denominator cannot be zero, as this would lead to an undefined expression. Similarly, if a function involves a square root, the expression under the root must be non-negative, as the square root of a negative number is not a real number. Logarithmic functions, on the other hand, are only defined for positive arguments. Therefore, when working with functions that combine these different elements, it's essential to consider all potential restrictions to accurately determine the domain. In essence, finding the domain is about identifying all the x-values that “play nicely” with the function and produce meaningful results. By carefully analyzing the function's structure and considering these restrictions, we can precisely define the set of permissible inputs. This understanding not only helps in graphing and analyzing functions but also in solving real-world problems where functions are used to model various phenomena.

The Function $f(x)=\frac{5x-4}{12x^2+17x-5}$

Let's consider the function $f(x)=\frac{5x-4}{12x^2+17x-5}$. This is a rational function, which means it is a ratio of two polynomials. The numerator is the linear expression 5x - 4, and the denominator is the quadratic expression 12x² + 17x - 5. When dealing with rational functions, the primary concern in determining the domain is identifying any values of x that would make the denominator equal to zero. Division by zero is undefined in mathematics, so any such values must be excluded from the domain. To find these values, we set the denominator equal to zero and solve for x. This involves solving the quadratic equation 12x² + 17x - 5 = 0. Quadratic equations can be solved using several methods, including factoring, completing the square, or using the quadratic formula. In this case, factoring is a straightforward approach. We look for two binomials that multiply to give the quadratic expression. The factors of 12x² are (4x) and (3x), and the factors of -5 are (-1) and (5) or (1) and (-5). Through trial and error, we find that the quadratic expression factors as (4x - 1)(3x + 5). Setting each factor equal to zero gives us the solutions 4x - 1 = 0 and 3x + 5 = 0. Solving these linear equations, we find x = 1/4 and x = -5/3. These are the values of x that make the denominator zero, and therefore, they must be excluded from the domain of the function. Thus, the domain of f(x) consists of all real numbers except x = 1/4 and x = -5/3. This means that any real number other than these two values can be input into the function, and it will produce a valid output.

Determining the Domain

The process of determining the domain for the function $f(x)=\frac{5x-4}{12x^2+17x-5}$ begins by recognizing it as a rational function. As rational functions involve division, our main concern is identifying any values of x that would make the denominator equal to zero, as division by zero is undefined. The denominator of our function is the quadratic expression 12x² + 17x - 5. To find the values of x that make this expression zero, we set the denominator equal to zero and solve the resulting quadratic equation: 12x² + 17x - 5 = 0. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. Factoring is often the quickest and most straightforward method when it is applicable. In this case, we look for two binomials that, when multiplied, give us the original quadratic expression. After some trial and error, we find that the quadratic expression can be factored as (4x - 1)(3x + 5). This means that 12x² + 17x - 5 is equivalent to (4x - 1)(3x + 5). Now, we set each factor equal to zero to find the values of x that make the entire expression zero: 4x - 1 = 0 and 3x + 5 = 0. Solving the first equation, 4x - 1 = 0, we add 1 to both sides to get 4x = 1, and then divide by 4 to find x = 1/4. Solving the second equation, 3x + 5 = 0, we subtract 5 from both sides to get 3x = -5, and then divide by 3 to find x = -5/3. These two values, x = 1/4 and x = -5/3, are the values that make the denominator of the function equal to zero. Therefore, they must be excluded from the domain of the function. The domain of f(x) includes all real numbers except these two values.

Solving the Quadratic Equation

To effectively solve the quadratic equation $12x^2 + 17x - 5 = 0$, several methods can be employed, each with its own advantages depending on the specific equation. One common approach is factoring, which involves expressing the quadratic expression as a product of two binomials. Factoring is often the quickest method when it's applicable, but it may not always be straightforward for more complex quadratics. In our case, the quadratic expression 12x² + 17x - 5 can be factored into (4x - 1)(3x + 5). This factorization is found by identifying two binomials that, when multiplied, yield the original quadratic expression. The factors of the leading term, 12x², are 4x and 3x, and the factors of the constant term, -5, are -1 and 5 or 1 and -5. Through a process of trial and error, we arrange these factors to achieve the correct middle term, 17x. Once the quadratic equation is factored, we set each factor equal to zero. This is based on the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we have 4x - 1 = 0 and 3x + 5 = 0. Solving each of these linear equations gives us the values of x that make the quadratic expression equal to zero. For the first equation, 4x - 1 = 0, we add 1 to both sides to get 4x = 1, and then divide by 4 to find x = 1/4. For the second equation, 3x + 5 = 0, we subtract 5 from both sides to get 3x = -5, and then divide by 3 to find x = -5/3. Therefore, the solutions to the quadratic equation 12x² + 17x - 5 = 0 are x = 1/4 and x = -5/3. These values are critical for determining the domain of the original function, as they represent the points where the denominator equals zero.

Expressing the Domain in Interval Notation

After identifying the values that are excluded from the domain, the next step is to express the domain in interval notation. Interval notation is a concise way to represent sets of real numbers using intervals. It employs parentheses and brackets to indicate whether the endpoints of an interval are included or excluded. Parentheses ( and ) are used to denote open intervals, where the endpoints are not included, while brackets [ and ] are used to denote closed intervals, where the endpoints are included. In our case, the domain of the function $f(x)=\frac{5x-4}{12x^2+17x-5}$ consists of all real numbers except x = 1/4 and x = -5/3. This means that the domain includes all numbers less than -5/3, all numbers between -5/3 and 1/4, and all numbers greater than 1/4. To express this in interval notation, we use three separate intervals. The first interval represents all numbers less than -5/3, which is written as (-∞, -5/3). The parenthesis on the left indicates that negative infinity is not included, as infinity is not a specific number but rather a concept. The parenthesis on the right indicates that -5/3 is not included in this interval, as it is a value that makes the denominator zero. The second interval represents all numbers between -5/3 and 1/4, which is written as (-5/3, 1/4). Both endpoints are excluded, as neither -5/3 nor 1/4 is in the domain. The third interval represents all numbers greater than 1/4, which is written as (1/4, ∞). The parenthesis on the left indicates that 1/4 is not included, and the parenthesis on the right indicates that positive infinity is not included. To combine these three intervals into a single expression for the domain, we use the union symbol, ∪. The union symbol means that we are combining all the numbers in the intervals into a single set. Therefore, the domain of the function f(x) in interval notation is (-∞, -5/3) ∪ (-5/3, 1/4) ∪ (1/4, ∞). This notation succinctly captures the idea that the domain includes all real numbers except -5/3 and 1/4.

Final Answer

In conclusion, the final answer for the domain of the function $f(x)=\frac{5x-4}{12x^2+17x-5}$ in interval notation is (-∞, -5/3) ∪ (-5/3, 1/4) ∪ (1/4, ∞). This solution is derived by first recognizing the function as a rational function, where the denominator cannot be equal to zero. We then set the denominator, 12x² + 17x - 5, equal to zero and solved the resulting quadratic equation. The solutions to the quadratic equation, x = -5/3 and x = 1/4, are the values that make the denominator zero and must be excluded from the domain. Finally, we expressed the domain in interval notation, which is a concise way to represent the set of all real numbers except for these two values. The interval notation (-∞, -5/3) represents all numbers less than -5/3, the interval (-5/3, 1/4) represents all numbers between -5/3 and 1/4, and the interval (1/4, ∞) represents all numbers greater than 1/4. The union symbol ∪ combines these three intervals to give the complete domain. This thorough process ensures that we have accurately identified all possible input values for the function that will produce a valid output, providing a comprehensive understanding of the function's behavior and limitations. Understanding the domain is crucial for various mathematical operations, including graphing, calculus, and solving real-world problems where functions are used to model different phenomena. Therefore, mastering the skill of determining the domain is an essential step in mathematical analysis.