Problem 15 - Intermediate Algebra

Question

Let \(f(x)=\frac{2 x+1}{x^{2}+3 x-4}\). Find a. \(f(0)\) b. \(f(2)\) c. \(f(1)\)

Step-by-Step Solution

Answer

a. \(-\frac{1}{4}\), b. \(\frac{5}{6}\), c. undefined

Step 1: Evaluate f(0)

To find \(f(0)\), substitute \(x = 0\) into the function: \[f(x) = \frac{2x + 1}{x^2 + 3x - 4}\]. This gives: \[f(0) = \frac{2(0) + 1}{0^2 + 3(0) - 4} = \frac{1}{-4} = -\frac{1}{4}.\]

Step 2: Evaluate f(2)

To find \(f(2)\), substitute \(x = 2\) into the function: \[f(x) = \frac{2x + 1}{x^2 + 3x - 4}\]. This gives: \[f(2) = \frac{2(2) + 1}{2^2 + 3(2) - 4} = \frac{4 + 1}{4 + 6 - 4} = \frac{5}{6}.\]

Step 3: Evaluate f(1)

To find \(f(1)\), substitute \(x = 1\) into the function: \[f(x) = \frac{2x + 1}{x^2 + 3x - 4}\]. This gives: \[f(1) = \frac{2(1) + 1}{1^2 + 3(1) - 4} = \frac{2 + 1}{1 + 3 - 4} = \frac{3}{0}.\] Since division by zero is undefined, \(f(1)\) is undefined.

Key Concepts

Function Evaluation

Function evaluation is the process by which we calculate the output of a function for particular values of its input variables. When dealing with mathematical functions, especially rational functions (those with a polynomial in the numerator and denominator), this process involves plugging a specific value into the given equation and simplifying it. For example, consider the function described as \( f(x) = \frac{2x + 1}{x^2 + 3x - 4} \). To evaluate this function at a point, say \( x = 0 \), you replace every occurrence of \( x \) in the equation with 0. This turns the function into a straightforward calculation:

• Substitute \( x = 0 \) into \( 2x + 1 \) to get \( 1 \).
• Replace \( x = 0 \) in \( x^2 + 3x - 4 \) to find \( -4 \).
• The result of \( f(0) \) after putting the numbers together is \( \frac{1}{-4} = -\frac{1}{4} \).

This example of function evaluation illustrates how to handle and simplify expressions, especially in rational functions.

Substitution Method

The substitution method is a straightforward technique used in evaluating functions and solving equations. This involves replacing the variable in the function or equation with a specific number. Substitution simplifies complex expressions into manageable calculations that provide the function's value at a certain point. Here's how it works with our given function and values:

• For \( f(x) = \frac{2x + 1}{x^2 + 3x - 4} \), when evaluating \( f(2) \), substitute 2 for every \( x \):
  • Numerator becomes \( 2(2) + 1 = 5 \).
  • Denominator becomes \( 2^2 + 3(2) - 4 = 6 \).
• This results in \( f(2) = \frac{5}{6} \).

This substitution simplifies the computation process, turning variable-based expressions into numeric calculations, thus helping understand potential behavior of the function at particular points.

Division by Zero

Division by zero is a concept where the denominator of a fraction becomes zero, making the entire expression undefined. In mathematics, any value divided by zero is not possible, leading to an undefined result. This principle holds utmost importance when evaluating rational functions.Consider the scenario of evaluating \( f(1) \) for the function \( f(x) = \frac{2x + 1}{x^2 + 3x - 4} \). Upon substitution, we find:

• Calculate the numerator: \( 2(1) + 1 = 3 \).
• Compute the denominator: \( 1^2 + 3(1) - 4 = 0 \).
• This results in \( f(1) = \frac{3}{0} \).

Since dividing by zero doesn't resolve into a meaningful or defined number, we conclude that \( f(1) \) is undefined. This highlights the importance of checking the denominator before performing division, ensuring rational function evaluations remain valid and meaningful.