Problem 10 - College Algebra and Calculus: An Applied Approach

Question

Find (a) \((f+g)(x)\), (b) \((f-g)(x)\), (c) \((f g)(x)\), and (d) \((f / g)(x)\). What is the domain of \(f / g\) ? \(f(x)=\sqrt{x^{2}-4}, \quad g(x)=\frac{x^{2}}{x^{2}+1}\)

Step-by-Step Solution

Answer

(a) \((f+g)(x) = \sqrt{x^{2}-4} + \frac{x^{2}}{x^{2}+1}\), (b) \((f-g)(x) = \sqrt{x^{2}-4} - \frac{x^{2}}{x^{2}+1}\), (c) \((fg)(x) = \sqrt{x^{2}-4} \cdot \frac{x^{2}}{x^{2}+1}\), (d) \((f / g)(x) = \frac{\sqrt{x^{2}-4}}{\frac{x^{2}}{x^{2}+1}}\). The domain of \(f / g\) is \(x <= -2\) or \(x >= 2\).

Step 1: Add the functions

To calculate \(f + g\), add \(f(x)\) and \(g(x)\) together. \[ (f+g)(x) = f(x) + g(x) = \sqrt{x^{2}-4} + \frac{x^{2}}{x^{2}+1} \]

Step 2: Subtract the functions

To calculate \(f - g\), subtract \(g(x)\) from \(f(x)\).\[ (f-g)(x) = f(x) - g(x) = \sqrt{x^{2}-4} - \frac{x^{2}}{x^{2}+1} \]

Step 3: Multiply the functions

To calculate \(f \cdot g\), multiply \(f(x)\) by \(g(x)\). \[ (fg)(x) = f(x) \cdot g(x) = \sqrt{x^{2}-4} \cdot \frac{x^{2}}{x^{2}+1} \]

Step 4: Divide the functions

To calculate \(f / g\), divide \(f(x)\) by \(g(x)\). Be aware that \(g(x)\) must not equal zero, as division by zero is undefined. \[ (f/g)(x) = f(x) / g(x) = \frac{\sqrt{x^{2}-4}}{\frac{x^{2}}{x^{2}+1}} \]

Step 5: Find the domain of \(f / g\)

The domain of \(f / g\) is all real numbers except for the values that make \(g(x) = 0\). In this case, \(g(x)\) is never zero, so the domain of \(f / g\) is all real numbers where both \(f(x)\) and \(g(x)\) are defined. Specifically, \(x^{2} - 4\) should be nonnegative for the expression under the square root to be real, which gives us \(x <= -2\) or \(x >= 2\). Combining these, the domain of \(f / g\) is \(x <= -2\) or \(x >= 2\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values, usually represented by \(x\), that allow the function to work without encountering undefined situations. For example, if a function involves a square root, like \(f(x) = \sqrt{x^2 - 4}\), we need the expression inside the square root to be non-negative (since the square root of a negative number isn't a real number). This means \(x^2 - 4 \geq 0\). Solving this inequality, we find that \(x \leq -2\) or \(x \geq 2\) are the acceptable values.

When dealing with functions like \(g(x) = \frac{x^2}{x^2 + 1}\), we must avoid division by zero. However, since \(x^2 + 1\) is always positive, \(g(x)\) is defined for all real numbers. The domain of \(f/g\), therefore, is affected by both \(f(x)\) and \(g(x)\), leading to the combination of these constraints, ultimately determining it as \(x \leq -2\) or \(x \geq 2\). Be mindful of these conditions, as they are crucial in function analysis.

Function Operations

Function operations include addition, subtraction, multiplication, and division of functions. These operations involve combining functions to create new ones. Here's a brief overview:

• **Addition**: The sum \((f+g)(x)\) is obtained by adding the outputs \(f(x)\) and \(g(x)\). For instance, \(f(x) + g(x) = \sqrt{x^2-4} + \frac{x^2}{x^2+1}\).
• **Subtraction**: Construct the difference \((f-g)(x)\) by subtracting \(g(x)\) from \(f(x)\). In the exercise, this looks like \(\sqrt{x^2-4} - \frac{x^2}{x^2+1}\).
• **Multiplication**: Get the product \((fg)(x)\) by multiplying \(f(x)\) and \(g(x)\), such as \(\sqrt{x^2-4} \cdot \frac{x^2}{x^2+1}\).
• **Division**: Obtain the quotient \((f/g)(x)\) by dividing \(f(x)\) by \(g(x)\). Ensure \(g(x)\) is never zero to avoid undefined outputs, yielding \(\frac{\sqrt{x^2-4}}{\frac{x^2}{x^2+1}}\).

These operations allow for intricate and dynamic function combinations, each producing distinct behavior and range based on the nature of \(f(x)\) and \(g(x)\). Always consider the domains, particularly for division, where care must be taken to avoid division by zero.

Rational Functions

Rational functions are defined as the ratio \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials. An important aspect of rational functions is ensuring the denominator \(Q(x)\) is not zero, as this would make the function undefined because division by zero is not possible.

For example, in an exercise where \(g(x) = \frac{x^2}{x^2 + 1}\), the denominator \(x^2 + 1\) is never zero since squaring \(x\) and adding one always results in a positive number. Thus, \(g(x)\) is defined for all \(x\).

Rational functions often entail detailed examination to determine their domain, places of discontinuity, and behavior at extreme (large positive or negative) values. They can exhibit complex behaviors like vertical asymptotes where the function is undefined or horizontal asymptotes indicating the end behavior as \(x\) approaches infinity or negative infinity. When combined with other function forms, like root functions, one must account for constraints from both components to determine the overall domain.