Problem 30 - Algebra A Combined Function

Question

If \(f(x)=\frac{x+8}{2 x-1}\) and \(g(x)=\frac{x-2}{x-5},\) find each function value. Find the domain of \(f(x)\).

Step-by-Step Solution

Answer

Domain of \(f(x)\) is all real numbers except \(x=\frac{1}{2}\).

Step 1: Identify Function Value of f(x)

Let's first consider the function \(f(x) = \frac{x+8}{2x-1}\) and find its expression for a particular \(x\). To find the function value, substitute a specific value of \(x\) into the function.

Step 2: Identify Function Value of g(x)

Now consider the function \(g(x) = \frac{x-2}{x-5}\) and similarly substitute a specific value of \(x\) into this function to find its respective value.

Step 3: Determine Domain of f(x)

The domain of \(f(x)\) is determined by its denominator. The expression \(2x-1\) should not be zero since division by zero is undefined. Solve the inequality \(2x-1eq0\).

Step 4: Solve for x in Domain Equation

To ensure the domain, solve \(2x-1=0\) to determine \(x\) values that are not included in the domain. Solving gives \(x=\frac{1}{2}\), so \(xeq\frac{1}{2}\) is the condition for the domain.

Key Concepts

Domain of a Function

When exploring functions, it's important to first understand the concept of "domain," which represents all the valid input values a function can accept. Specifically, the domain is all the numbers you can substitute for the variable without making something mathematically impossible, like dividing by zero or taking the square root of a negative number in real numbers.

For rational functions, the domain can present specific challenges. Take for instance the function given as: \[ f(x) = \frac{x+8}{2x-1} \]This is a rational function because it involves a ratio (or fraction) with a polynomial in the denominator. To find the domain of \( f(x) \), you must ensure the denominator doesn't equal zero because division by zero is undefined. This involves solving the equation given by setting the denominator to zero: \[ 2x - 1 = 0 \]

Solving for \( x \) gives the result \( x = \frac{1}{2} \), meaning this specific value is everything the domain isn't—it is where the function "breaks." Thus, the domain of \( f(x) \) is all real numbers except \( \frac{1}{2} \). In an interval notation, you would write this as:

• \( (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty) \)

By eliminating points where the function behaves erratically, you can be sure your function behaves properly!

Rational Functions

Rational functions are fascinating because they are formed as the ratio of two polynomials. Generally speaking, a rational function looks like this:\[ f(x) = \frac{P(x)}{Q(x)} \]where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). In such functions, the behavior of the function is heavily determined by the polynomial in the denominator.

This is why understanding rational functions involves diligently checking where the denominator becomes zero. In our exercise, consider \( f(x) = \frac{x+8}{2x-1} \) and \( g(x) = \frac{x-2}{x-5} \). Both are examples of rational functions:

• The denominator of \( f(x) \), \( 2x-1 \), leads to a division by zero when \( x = \frac{1}{2} \).
• The denominator of \( g(x) \), \( x-5 \), leads to a division by zero when \( x = 5 \).

By knowing these restrictions, you can better comprehend how these functions act and where any discontinuities or undefined points are. Handling these aspects properly is a major part of working with rational functions effectively.

Function Evaluation

Function evaluation is a fundamental skill in algebra. This process involves substituting a given number for the variable in the function's formula. Doing so allows you to compute the output value for that particular input. Function evaluation is crucial for exploring how the function behaves at specific points.

Using our example functions:

• For \( f(x) = \frac{x+8}{2x-1} \), to evaluate it at \( x = 3 \), you'd substitute \( 3 \) for \( x \) into the function, leading to:\[ f(3) = \frac{3+8}{2(3)-1} = \frac{11}{5} \]
• Similarly for \( g(x) = \frac{x-2}{x-5} \), if you evaluate it at \( x = 10 \), you substitute \( 10 \):\[ g(10) = \frac{10-2}{10-5} = \frac{8}{5} \]

The steps are straightforward: substitute, simplify, and solve. Function evaluation guides you to understand the outputs for inputs you're interested in, which is particularly helpful in more extensive applications like graphing the function or solving real-world problems. By mastering evaluation, complex problems become more approachable!