Problem 13 - Beginning and Intermediate Algebra

Question

For each pair of functions, find a) \(\left(\frac{f}{g}\right)(x)\) and b \(\left(\frac{f}{g}\right)(-2)\) Identify any values that are not in the domain of \(\left(\frac{f}{g}\right)(x)\). $$f(x)=3 x^{2}+14 x+8, g(x)=3 x+2$$

Step-by-Step Solution

Answer

So, the short answer is: a) \(\frac{f}{g}(x) = \frac{3 x^{2} + 14 x + 8}{3 x + 2}\) b) \(\frac{f}{g}(-2) = 2\) The domain of \(\frac{f}{g}(x)\) excludes \(x = -\frac{2}{3}\).

Step 1: Find the quotient of the functions

First, we'll find the quotient \(\frac{f}{g}(x)\), which is the result of dividing \(f(x)\) by \(g(x)\): \[\frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{3 x^{2} + 14 x + 8}{3 x + 2}\]

Step 2: Simplify the quotient if possible

If possible, try to simplify the quotient. In this case, we can use polynomial long division or synthetic division to check if there is any simplification. Since there is no clear common factor between the numerator and denominator, we can assume the expression is in its simplest form: \[\frac{f}{g}(x) = \frac{3 x^{2} + 14 x + 8}{3 x + 2}\]

Step 3: Evaluate the quotient at the specific value

Now, we'll find \(\frac{f}{g}(-2)\) by plugging in \(x=-2\) into our quotient: \[\frac{f}{g}(-2) = \frac{3(-2)^2 + 14(-2) + 8}{3(-2) + 2} = \frac{12 - 28 + 8}{-6 +2}\] Simplify the expression: \[\frac{f}{g}(-2) = \frac{-8}{-4} = 2\]

Step 4: Identify any values not in the domain of the quotient function

The domain of the quotient function will exclude any values of \(x\) for which the denominator is equal to zero. So we need to find any values of \(x\) for which \(g(x)=0\): \[3x + 2 = 0\] Solve for \(x\): \[x = -\frac{2}{3}\] Therefore, the domain of \(\frac{f}{g}(x)\) excludes \(-\frac{2}{3}\). #Summary# 1. We found the quotient of the functions: \(\frac{f}{g}(x) = \frac{3 x^{2} + 14 x + 8}{3 x + 2}\) 2. We simplified the expression. 3. We evaluated the quotient at \(x=-2\): \(\frac{f}{g}(-2) = 2\) 4. We identified the domain of the quotient function excludes: \(x = -\frac{2}{3}\)

Key Concepts

Domain of a Function

The domain of a function refers to all the values that can be plugged into the function without causing any problems like division by zero or square roots of negative numbers.
In the context of rational functions, which are ratios of polynomials, the domain is all real numbers except where the denominator is zero.
For example, if we have a rational function \( \frac{f(x)}{g(x)} \), its domain is determined by the values of \( x \) where \( g(x) eq 0 \).

To find the domain of \( \frac{f}{g}(x) \), first, set the denominator equal to zero:

• Find \( x \) values where \( g(x) = 0 \).
• Exclude these \( x \) values from the domain.

In our problem, the denominator is \( g(x) = 3x + 2 \). Set it to zero: \( 3x + 2 = 0 \).
Solving \( x = -\frac{2}{3} \).
So, the function \( \frac{f}{g}(x) \) is not defined at \( x = -\frac{2}{3} \).
Thus, the domain of \( \frac{f}{g}(x) \) is all real numbers except \( -\frac{2}{3} \).

Polynomial Long Division

Polynomial long division is a method used to divide polynomials, similar to the long division process for numbers.
It's helpful for breaking down complex fractions in rational expressions into a simpler form.

Here's a simplified process:

• Divide the first term of the dividend by the first term of the divisor.
• Multiply the entire divisor by the result and subtract from the dividend.
• Repeat the process with the new polynomial that you get as a result.

For example, to divide \( 3x^2 + 14x + 8 \) by \( 3x + 2 \), you would:

1. Divide the leading term \( 3x^2 \) by \( 3x \) to get \( x \). Multiply \( x \) by \( 3x + 2 \) to get \( 3x^2 + 2x \).
2. Subtract \( 3x^2 + 2x \) from \( 3x^2 + 14x + 8 \) to get \( 12x + 8 \).
3. Divide \( 12x \) by \( 3x \), which gives \( 4 \), then multiply \( 4 \) by \( 3x + 2 \) to get \( 12x + 8 \).
4. Subtract \( 12x + 8 \) from \( 12x + 8 \), which leaves \( 0 \).

The quotient is \( x + 4 \) and any remainder would be included over the original divisor. This clean division means there's no remainder here.

Synthetic Division

Synthetic division is a simpler, quicker method compared to polynomial long division, especially when dividing by a linear expression of the form \( x - c \).
It's an efficient process for dividing a polynomial by a binomial.Here is how it works:

• Use the root \( c \) from \( x - c \) which the polynomial is divided by.
• Arrange the coefficients of the polynomial in a row.
• Bring down the first coefficient.
• Multiply it by \( c \) and add it to the next coefficient.
• Repeat the multiplication and addition until you reach the end.
• The last number will be the remainder, and the other numbers form the quotient.

Using synthetic division is easier to handle because it involves less notation and fewer subtraction operations than long division. It's particularly useful for determining real roots quickly.

Function Evaluation

Evaluating a function involves finding the output value of a function for a specific input value.
It is a straightforward process that requires replacing the variable in the function with the given number.
For rational functions, this means taking the given \( x \) value and plugging it into both the numerator and the denominator if performed from the quotient form. To evaluate \( \frac{f}{g}(-2) \), you substitute \( -2 \) into the functions \( f(x) \) and \( g(x) \):

• Plug \( x = -2 \) into \( f(x) = 3x^2 + 14x + 8 \) which results in \( 3(-2)^2 + 14(-2) + 8 = 12 - 28 + 8 = -8 \).
• Substitute \( x = -2 \) into \( g(x) = 3x + 2 \) resulting in \( 3(-2) + 2 = -6 + 2 = -4 \).

Finally, divide the results: \( \frac{-8}{-4} = 2 \).
This process determines the specific output of the rational function for that particular input value.