Problem 19 - College Algebra with Modeling and Visualization

Question

Exercises \(11-30:\) Use \(f(x)\) and \(g(x)\) to find a formula for each expression. Identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d) \((f / g)(x)\) $$ f(x)=\frac{1}{x+1}, \quad g(x)=\frac{3}{x+1} $$

Step-by-Step Solution

Answer

Formulas: \(\frac{4}{x+1}, \frac{-2}{x+1}, \frac{3}{(x+1)^2}, \frac{1}{3}\); Domain: \(x \neq -1\).

Step 1: Understanding the Problem

We have two functions, \(f(x) = \frac{1}{x+1}\) and \(g(x) = \frac{3}{x+1}\). The task is to find the resulting formula for \((f+g)(x)\), \((f-g)(x)\), \((fg)(x)\), and \((f/g)(x)\) and determine the domain for each expression.

Step 2: Calculate (f + g)(x)

To find \((f+g)(x)\), we add up the functions:\[(f+g)(x) = f(x) + g(x) = \frac{1}{x+1} + \frac{3}{x+1} = \frac{1 + 3}{x+1} = \frac{4}{x+1}\]The denominator \((x+1)\) cannot be zero, so the domain excludes \(x = -1\). Thus, the domain is \(x \in \mathbb{R} \setminus \{-1\}\).

Step 3: Calculate (f - g)(x)

To find \((f-g)(x)\), we subtract the functions:\[(f-g)(x) = f(x) - g(x) = \frac{1}{x+1} - \frac{3}{x+1} = \frac{1 - 3}{x+1} = \frac{-2}{x+1}\]The denominator \((x+1)\) cannot be zero, thus the domain is \(x \in \mathbb{R} \setminus \{-1\}\).

Step 4: Calculate (f g)(x)

To find \((fg)(x)\), we multiply the functions:\[(fg)(x) = f(x) \cdot g(x) = \frac{1}{x+1} \times \frac{3}{x+1} = \frac{3}{(x+1)^2}\]The denominator \((x+1)^2\) cannot be zero, giving the domain \(x \in \mathbb{R} \setminus \{-1\}\).

Step 5: Calculate (f / g)(x)

To find \((f/g)(x)\), we divide the functions:\[(f/g)(x) = \frac{f(x)}{g(x)} = \frac{\frac{1}{x+1}}{\frac{3}{x+1}} = \frac{1}{x+1} \cdot \frac{x+1}{3} = \frac{1}{3}\]No additional constraints from \(x+1\) simplifying out, but originally, none can be zero, hence the domain remains \(x \in \mathbb{R} \setminus \{-1\}\).

Step 6: Conclusion

All expressions for the given combinations are determined. Domains for \((f + g)(x)\), \((f - g)(x)\), \((fg)(x)\), and \((f/g)(x)\) exclude \(x = -1\), so the domain is \(x \in \mathbb{R} \setminus \{-1\}\).

Key Concepts

Domain of a Function

The domain of a function is crucial in understanding where the function can operate. In simple terms, it's a set of all possible input values (usually represented by \(x\)) that will not cause any mathematical errors, such as division by zero or taking square roots of negative numbers. For rational functions like the ones in this exercise, identifying the domain involves ensuring that the denominator isn't equal to zero. This is because division by zero is undefined and a critical value, which must be excluded from the domain. For example, in the functions \(f(x) = \frac{1}{x+1}\) and \(g(x) = \frac{3}{x+1}\), the denominator changes to zero when \(x = -1\). Hence, to avoid division by zero, the domain excludes \(x = -1\), making it \(x \in \mathbb{R} \setminus \{-1\}\). Remember:

• Examine the denominator.
• Set the denominator not equal to zero and solve for \(x\).
• Exclude these critical values from the domain.

Understanding and determining domains are essential for working with functions, especially when combining them using operations like addition, subtraction, multiplication, and division.

Algebraic Expressions

Algebraic expressions are combinations of constants, variables, and algebraic operations (like addition and multiplication). Each term in an expression can stand alone to convey a part of the solution or problem. For example, in the exercise functions \(f(x) = \frac{1}{x+1}\) and \(g(x) = \frac{3}{x+1}\), both are fractional expressions that show clear algebraic relations. When performing operations on these, such as

• Adding - \((f+g)(x) = \frac{1}{x+1} + \frac{3}{x+1} = \frac{4}{x+1}\)
• Subtracting - \((f-g)(x) = \frac{1}{x+1} - \frac{3}{x+1} = \frac{-2}{x+1}\)

different rules apply to simplify them. The key to handling algebraic expressions is recognizing each part (like numerators and denominators in fractions), knowing how to combine like terms, and applying operations correctly. Simplification often involves canceling out common factors or terms. Keep practicing these operations to build a solid foundation in algebraic understanding!

Rational Functions

Rational functions are essentially fractions where both the numerator and the denominator are algebraic expressions. For a function to be rational, its form must be \(\frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are polynomials. In the exercise, we deal with functions like \(f(x) = \frac{1}{x+1}\), a straightforward example of a rational function. Rational functions can be combined using mathematical operations, such as:

• Multiplication: For example, \((fg)(x) = \frac{1}{x+1} \times \frac{3}{x+1} = \frac{3}{(x+1)^2}\), where both numerators and denominators multiply together.
• Division: Here, \((f/g)(x) = \frac{1}{3}\), which simplifies as the terms in the numerator and denominator can cancel each other out.

Understanding rational functions involves recognizing how operations and simplifications like factoring or expanding affect the overall function. Always remember, rational functions are defined as long as their denominators are non-zero, an important point in both determining their domain and conducting any algebraic operation.