Problem 5 - College Algebra

Question

For the following exercises, determine the domain for each function in interval notation. Given \(f(x)=x^{2}+2 x\) and \(g(x)=6-x^{2},\) find \(f+g\) \(f-g, f g,\) and \(\frac{f}{g}\)

Step-by-Step Solution

Answer

The domain for \( f+g, f-g, fg \) is \((-\infty, \infty)\); for \( \frac{f}{g} \), it's \((-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty)\).

Step 1: Define Functions

We start by recognizing the given functions. Function \( f(x) = x^2 + 2x \) and function \( g(x) = 6 - x^2 \). These are polynomial functions.

Step 2: Determine Domain of Functions

Since both \( f(x) \) and \( g(x) \) are polynomial functions, each is defined for all real numbers. Therefore, the domain for both \( f(x) \) and \( g(x) \) is \( (-\infty, \infty) \).

Step 3: Sum of Functions (f+g)

To find \( (f+g)(x) \), we add the two functions: \( f(x) + g(x) = (x^2 + 2x) + (6 - x^2) = 2x + 6 \). This is a linear function which is also defined for all real numbers. The domain is \( (-\infty, \infty) \).

Step 4: Difference of Functions (f-g)

To find \( (f-g)(x) \), subtract the second function from the first: \( f(x) - g(x) = (x^2 + 2x) - (6 - x^2) = 2x^2 + 2x - 6 \). This is a quadratic function with the domain \( (-\infty, \infty) \).

Step 5: Product of Functions (fg)

To find \( (fg)(x) \), multiply the functions: \( f(x) \cdot g(x) = (x^2 + 2x)(6 - x^2) = 6x^2 + 12x - x^4 - 2x^3 \). This product yields a polynomial function, also defined for all real numbers. The domain is \( (-\infty, \infty) \).

Step 6: Quotient of Functions (f/g)

To find \( \frac{f}{g}(x) \), we take \( f(x) \) and divide by \( g(x) \): \( \frac{x^2 + 2x}{6 - x^2} \). The domain is restricted where \( g(x) eq 0 \), which is where \( 6 - x^2 eq 0 \). Solving \( 6 - x^2 = 0 \) gives \( x = \pm\sqrt{6} \). Thus, the domain in interval notation is \( (-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty) \).

Key Concepts

Domain of Functions

When dealing with functions, the domain represents all the possible input values (usually denoted as "x") that will produce a valid output. In simple terms, the domain includes all real numbers for which the function is defined. For polynomial functions like the ones given in the exercise, which include expressions involving powers of x, the domain is typically all real numbers. This is because polynomials are well-defined for any real input.

For instance, both functions, \( f(x) = x^2 + 2x \) and \( g(x) = 6 - x^2 \), are polynomials. Neither of these functions involves any division by zero or square roots of negative numbers, so their domains span all real numbers. In mathematical terms, we express their domain as \((-\infty, \infty)\) in interval notation, meaning x can be any real number.

Interval Notation

Interval notation is a method of writing down the set of numbers that form the domain of a function. It's a concise way to describe the range of inputs using intervals.

In interval notation:

• A parenthesis \(()\) is used when an endpoint is not included in the interval.
• A bracket \([]\) is used when an endpoint is included.
• The symbol \((-)\infty \) denotes that there is no bound on the interval in that direction, but it's always accompanied by a parenthesis because infinity is not a specific number we can actually reach or include.

For \( f(x) \) and \( g(x) \), which are polynomial functions, the domain, expressed in interval notation, is \((-\infty, \infty)\). This is because every real number is an acceptable input.

However, when it comes to the function division like \( \frac{f}{g} \), we need to exclude values that make the denominator zero. In such cases, interval notation skillfully represents the domain excluding those points, like \((-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty)\), where \(x = \pm \sqrt{6}\) are not included.

Function Addition and Subtraction

When you add or subtract functions, you combine their outputs. The input remains the same, but the output value is altered by either adding or subtracting each function's result.

For example, if you're given two functions, \( f(x) \) and \( g(x) \), and asked to find \( f + g \), you simply add them: \( (f + g)(x) = f(x) + g(x) \). Similarly, for subtraction, it follows \( (f - g)(x) = f(x) - g(x) \).

In the exercise, the sum \( (f+g)(x) = 2x + 6 \) simplifies to a linear function, while the subtraction \( (f-g)(x) = 2x^2 + 2x - 6 \) results in a quadratic function. Both operations preserve the domain of \((-\infty, \infty)\) as long as no restrictions are introduced like denominators becoming zero, which is not the case here.

Function Multiplication and Division

Function multiplication involves multiplying the outputs of two functions. For functions \( f(x) \) and \( g(x) \), the product function is defined as \( (fg)(x) = f(x) \cdot g(x) \).

Function division, on the other hand, requires more caution. When dividing \( f(x) \) by \( g(x) \), the result is \( \frac{f}{g}(x) = \frac{f(x)}{g(x)} \). Here, it's crucial that \( g(x) eq 0 \) to avoid undefined expressions.

For the provided functions in the exercise, the multiplication \( (fg)(x) \) creates another polynomial \( 6x^2 + 12x - x^4 - 2x^3 \), which retains the domain of all real numbers \((-\infty, \infty)\).

However, the division \( \frac{f}{g} \) requires ensuring \( g(x) \) does not equal zero, which introduces restrictions at \( x = \pm \sqrt{6} \), so its domain becomes \( (-\infty, -\sqrt{6}) \cup (-\sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, \infty) \). This emphasizes the need to investigate the denominator when performing division.