Problem 12 - 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) \((f / g)(x)\) $$ f(x)=1-4 x, \quad g(x)=3 x+1 $$

Step-by-Step Solution

Answer

(a) 2-x with domain all reals; (b) -7x with domain all reals; (c) -12x^2-x+1 with domain all reals; (d) \(\frac{1-4x}{3x+1}\) with domain all reals except \(x=-\frac{1}{3}\).

Step 1: Finding (f+g)(x)

To determine the formula for \((f+g)(x)\), add the two functions together: \[ (f+g)(x) = f(x) + g(x) = (1-4x) + (3x+1) \] Simplify the expression: \[ (f+g)(x) = 1 - 4x + 3x + 1 = 2 - x \] The domain of \((f+g)(x)\) is all real numbers since there are no restrictions such as division by zero or square roots of negative numbers.

Step 2: Finding (f-g)(x)

Subtract the function \(g(x)\) from \(f(x)\) to obtain \((f-g)(x)\): \[ (f-g)(x) = f(x) - g(x) = (1-4x) - (3x+1) \] Simplify the expression: \[ (f-g)(x) = 1 - 4x - 3x - 1 = -7x \] The domain of \((f-g)(x)\) is all real numbers due to the absence of restrictions on \(x\).

Step 3: Finding (f * g)(x)

Multiply the functions \(f(x)\) and \(g(x)\) to get \((f \cdot g)(x)\): \[ (f \cdot g)(x) = f(x) \cdot g(x) = (1-4x)(3x+1) \] Use the distributive property to simplify: \[ (f \cdot g)(x) = 1 \cdot 3x + 1 \cdot 1 - 4x \cdot 3x - 4x \cdot 1 = 3x + 1 - 12x^2 - 4x \] Combine like terms: \[ (f \cdot g)(x) = -12x^2 - x + 1 \] The domain for \((f \cdot g)(x)\) is all real numbers, as multiplication introduces no new constraints on the domain.

Step 4: Finding (f / g)(x)

Divide \(f(x)\) by \(g(x)\) to determine \((f / g)(x)\): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{1-4x}{3x+1} \] The expression is defined for all \(x\) such that \(g(x) eq 0\). Solve \(3x + 1 = 0\) for \(x\): \[ 3x + 1 = 0 \rightarrow 3x = -1 \rightarrow x = -\frac{1}{3} \] Thus, the domain of \((f / g)(x)\) is all real numbers excluding \(x = -\frac{1}{3}\).

Key Concepts

Function Addition

In College Algebra, adding functions is as simple as combining like terms when you add the expressions that make up the functions. If you have two functions, say \( f(x) \) and \( g(x) \), the function addition is denoted as \((f + g)(x)\). In our example,

• f(x) = 1 - 4x
• g(x) = 3x + 1

To find \((f+g)(x)\), add the two functions:\[(f+g)(x) = f(x) + g(x) = (1 - 4x) + (3x + 1)\] Combine the like terms to simplify:\[(f+g)(x) = 1 - 4x + 3x + 1 = 2 - x\] The formula for \((f+g)(x)\) is \(2-x\) and its domain is all real numbers. This is because there are no calculations within the function that would restrict values of \(x\), such as square roots or denominators in fractions.

Function Subtraction

Subtracting functions is very similar to adding them. Here, we think of it as simply taking the difference of the two function expressions. Given functions \(f(x)\) and \(g(x)\), the subtraction is represented as \((f - g)(x)\). If we look at what we have:

• f(x) = 1 - 4x
• g(x) = 3x + 1

The subtraction process involves:\[(f-g)(x) = f(x) - g(x) = (1-4x) - (3x+1)\] Make sure to distribute the negative sign across the \(g(x)\) function and combine like terms:\[(f-g)(x) = 1 - 4x - 3x - 1 = -7x\] Thus, the resulting expression is \(-7x\), with the domain remaining as all real numbers since there are no limiting operations such as division that would exclude any values of \(x\).

Function Multiplication

Multiplying functions entails applying the distributive property, which ensures each term in one function multiplies with every term in the other. When you multiply \( f(x) \) and \( g(x) \), it is represented as \((f \cdot g)(x)\). Let's consider:

• f(x) = 1 - 4x
• g(x) = 3x + 1

To multiply these functions, use the following steps:\[(f \cdot g)(x) = (1-4x)(3x+1)\]Expand using distributive property:\[(f \cdot g)(x) = 1 \cdot 3x + 1 \cdot 1 - 4x \cdot 3x - 4x \cdot 1\]Simplify the expression:\[(f \cdot g)(x) = 3x + 1 - 12x^2 - 4x \]Combine like terms to obtain:\[(f \cdot g)(x) = -12x^2 - x + 1\]The domain of the resulting expression is all real numbers, as there is no division, which might otherwise restrict certain values of \(x\).

Function Division

Division of functions involves dividing the expression of one function by the other, represented as \((\frac{f}{g})(x)\). However, precautions must be taken to ensure the denominator does not equal zero, as division by zero is undefined.For our functions:

• f(x) = 1 - 4x
• g(x) = 3x + 1

Perform the division:\[\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{1-4x}{3x+1}\]To find the domain, eliminate values of \(x\) that make \(g(x)\) zero:- Solve \(3x + 1 = 0\) to find when \(g(x) = 0\):\[3x + 1 = 0 \rightarrow 3x = -1 \rightarrow x = -\frac{1}{3}\]Hence, the domain for \((\frac{f}{g})(x)\) excludes \(x = -\frac{1}{3}\). This means the valid domain for the expression is all real numbers except \(x = -\frac{1}{3}\).