Problem 12 - 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 \(\left(\frac{f}{g}\right)(x)\). $$f(x)=x^{2}-15 x+54, g(x)=x-9$$

Step-by-Step Solution

Answer

a) \(\left(\frac{f}{g}\right)(x) = \frac{x^{2}-15 x+54}{x-9}\), b) \(\left(\frac{f}{g}\right)(-2) = -8\), and the value not in the domain of \(\left(\frac{f}{g}\right)(x)\) is x = 9.

Step 1: Find \(\left(\frac{f}{g}\right)(x)\)

To find \(\left(\frac{f}{g}\right)(x)\), we will simply divide f(x) by g(x). Given functions are: \( f(x)=x^{2}-15 x+54\) \(g(x)=x-9\) So, \(\left(\frac{f}{g}\right)(x)\) = \(\frac{f(x)}{g(x)}\) = \(\frac{x^{2}-15 x+54}{x-9}\)

Step 2: Calculate \(\left(\frac{f}{g}\right)(-2)\)

Now, let's find \(\left(\frac{f}{g}\right)(-2)\) by substituting x = -2 into the expression obtained in step 1. \(\left(\frac{f}{g}\right)(-2)\) = \(\frac{(-2)^{2}-15 (-2)+54}{(-2)-9}\) \(\left(\frac{f}{g}\right)(-2)\) = \(\frac{4 + 30 + 54}{-11}\) \(\left(\frac{f}{g}\right)(-2)\) = \(\frac{88}{-11}\) \(\left(\frac{f}{g}\right)(-2)\) = -8 Now we have the value of \(\left(\frac{f}{g}\right)(-2)\) which is -8.

Step 3: Identify the values not in the domain of \(\left(\frac{f}{g}\right)(x)\)

To find the values not in the domain of \(\left(\frac{f}{g}\right)(x)\), we need to identify the values at which g(x) becomes zero, because division by zero is undefined in mathematical expressions. So, let's find the value of x at which g(x) = 0. \(g(x) = x-9 = 0\) \(x = 9\) x = 9 is the value at which g(x) = 0, so it is not in the domain of \(\left(\frac{f}{g}\right)(x)\). In conclusion, a) \(\left(\frac{f}{g}\right)(x) = \frac{x^{2}-15 x+54}{x-9}\), b) \(\left(\frac{f}{g}\right)(-2) = -8\), and the value that is not in the domain of \(\left(\frac{f}{g}\right)(x)\) is x = 9.

Key Concepts

Function Division

Function division is a fundamental concept in mathematics. It involves dividing one function by another. In our exercise, we have two functions:

• \( f(x) = x^2 - 15x + 54 \)
• \( g(x) = x - 9 \)

To find the division \( \left( \frac{f}{g} \right)(x) \), you essentially form a new function by dividing the first function by the second. It's pivotal to write this as a fraction:\[\left( \frac{f}{g} \right)(x) = \frac{x^2 - 15x + 54}{x - 9}\]The numerator and the denominator are polynomials here. One critical task in function division is checking whether the denominator becomes zero. This is because division by zero is undefined and can lead to discontinuities in the function.

Domain Restriction

Domain restriction happens when certain values of \( x \) make the denominator of a fractional expression zero. These are the values that we need to exclude from the domain of the rational function.For function \( g(x) = x - 9 \), we find where it becomes zero.

• Set \( g(x) = 0 \), leading to \( x - 9 = 0 \).
• Solve the equation; we find \( x = 9 \).

Since the denominator becomes zero at \( x = 9 \), \( x = 9 \) is not part of the domain of \( \left( \frac{f}{g} \right)(x) \). This is a common process in rational functions to avoid undefined points.
Thus, the domain of \( \left( \frac{f}{g} \right)(x) \) is all real numbers except \( x = 9 \). Excluding these critical values from the domain ensures that the function remains continuous and properly defined everywhere else.

Evaluating Functions

Evaluating functions involves substituting a specific value for \( x \) in the function expression and calculating the result. Let's see how this works with our function \( \left( \frac{f}{g} \right)(x) \).In our exercise, we need to find \( \left( \frac{f}{g} \right)(-2) \). You substitute \( -2 \) into both the numerator and the denominator:

• \((-2)^2 - 15(-2) + 54\) for the numerator.
• \((-2) - 9\) for the denominator.

So, the calculation is:\[\left( \frac{f}{g} \right)(-2) = \frac{4 + 30 + 54}{-11} = \frac{88}{-11} = -8\]By substituting and simplifying, we find that \( \left( \frac{f}{g} \right)(-2) \) is \(-8\). Evaluating functions like this allows us to find specific points on the function's graph or real-world interpretations when such functions model applicable scenarios.