Problem 9 - 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 \(\left(\frac{f}{g}\right)(x)\) $$f(x)=6 x+9, g(x)=x+4$$

Step-by-Step Solution

Answer

a) \(\left(\frac{f}{g}\right)(x) = \frac{6x + 9}{x + 4}\) b) \(\left(\frac{f}{g}\right)(-2) = \frac{-3}{2}\) The function is undefined at \(x = -4\).

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

To find the function \(\left(\frac{f}{g}\right)(x)\), we need to divide the function \(f(x)\) by the function \(g(x)\). Given that \(f(x) = 6x + 9\) and \(g(x) = x + 4\), we get: \[\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{6x + 9}{x + 4}\]

Step 2: Identify values not in the domain

The function \(\left(\frac{f}{g}\right)(x)\) will be undefined for any values of x that cause the denominator to be zero. To find these values, we set the denominator equal to zero and solve for x: \(g(x) = x + 4 = 0\) Solving for x, we get: \(x = -4\) So, the function \(\left(\frac{f}{g}\right)(x)\) is undefined at \(x = -4\).

Step 3: Evaluate the function at \(x = -2\)

To find \(\left(\frac{f}{g}\right)(-2)\), substitute \(x = -2\) into the function \(\left(\frac{f}{g}\right)(x)\) from Step 1: \[\left(\frac{f}{g}\right)(-2) = \frac{6(-2) + 9}{(-2) + 4}\] Simplify the expression: \[\left(\frac{f}{g}\right)(-2) = \frac{-12+ 9}{2} = \frac{-3}{2}\] The final results are: a) \(\left(\frac{f}{g}\right)(x) = \frac{6x + 9}{x + 4}\) b) \(\left(\frac{f}{g}\right)(-2) = \frac{-3}{2}\) The function \(\left(\frac{f}{g}\right)(x)\) is undefined at \(x = -4\).

Key Concepts

Division of Functions

Function division involves creating a new function by dividing one function by another. When we have two functions, like \( f(x) = 6x + 9 \) and \( g(x) = x + 4 \), we can form a division of functions, denoted as \( \left(\frac{f}{g}\right)(x) \). This operation means taking the output of \( f(x) \), which is \( 6x + 9 \), and dividing it by the output of \( g(x) \), which is \( x + 4 \). Hence, the division results in:

• \( \left(\frac{f}{g}\right)(x) = \frac{6x + 9}{x + 4} \)

Creating a division function is a simple process but be aware that it might introduce constraints to the original functions. We need to be cautious about values that make the denominator zero, as these will impact the domain.

Domain of a Function

The domain of a function is all the possible inputs (\( x \) values) for which the function is defined. When dealing with division of functions, we must consider restrictions that result from the division. Specifically, the denominator must not be zero, as division by zero is undefined. In our example of \( \left(\frac{f}{g}\right)(x) = \frac{6x + 9}{x+4} \), the critical task involves finding values of \( x \) that make \( g(x) = x + 4 = 0 \). By solving \( x + 4 = 0 \), we determine:

• \( x = -4 \)

This value cannot be included in the domain of \( \left(\frac{f}{g}\right)(x) \), as it results in division by zero. Therefore, the domain of the function \( \left(\frac{f}{g}\right)(x) \) is all real numbers except \( x = -4 \).

Function Evaluation

Function evaluation simplifies the process of finding the output of the function for a specific input value. When we evaluate \( \left( \frac{f}{g} \right)(x) \) at a specific point, say \( x = -2 \), we substitute \( x \) with \( -2 \):\[ \left( \frac{f}{g} \right)(-2) = \frac{6(-2) + 9}{-2 + 4} \]Through calculation, this expression simplifies as:

• Firstly, solve the numerator: \( 6(-2) + 9 = -12 + 9 = -3 \)
• Secondly, solve the denominator: \(-2 + 4 = 2 \)
• Resulting in: \( \frac{-3}{2} \)

Thus, \( \left( \frac{f}{g} \right)(-2) = \frac{-3}{2} \). Function evaluation allows you to see how the function behaves at specific points by simply plugging in the value into the expression. Remember always to ensure the input value is in the function's domain before evaluating.