Problem 21 - A Graphical Approach to College Algebra

Question

Let \(f(x)=x^{2}+3 x\) and \(g(x)=2 x-1 .\) Perform the composition or operation indicated. $$\left(\frac{g}{f}\right)(0)$$

Step-by-Step Solution

Answer

The composition \( \left(\frac{g}{f}\right)(0) \) is undefined due to division by zero.

Step 1: Understand the Composition

First, let’s understand what is being asked. We are asked to find \( \left(\frac{g}{f}\right)(0) \). This means we need to evaluate the composition \( \frac{g(x)}{f(x)} \) at \( x = 0 \).

Step 2: Evaluate g(x) at x=0

We need to find the value of \( g(0) \). Given that \( g(x) = 2x - 1 \), we plug in \( x = 0 \):\[g(0) = 2(0) - 1 = -1.\]

Step 3: Evaluate f(x) at x=0

Next, find the value of \( f(0) \). Given that \( f(x) = x^2 + 3x \), we substitute \( x = 0 \):\[f(0) = 0^2 + 3(0) = 0.\]

Step 4: Determine the Composition

Finally, substitute these values into the expression \( \frac{g(x)}{f(x)} \):\[\left(\frac{g}{f}\right)(0) = \frac{g(0)}{f(0)} = \frac{-1}{0}.\]Here we see the fraction is undefined because division by zero is not allowed.

Key Concepts

Rational Functions

A rational function is an expression that can be written as a fraction where both the numerator and the denominator are polynomials. Simply put, it's like having one algebraic expression divided by another. The general form of a rational function looks like this: \[ \frac{p(x)}{q(x)} \]where both \(p(x)\) and \(q(x)\) are polynomials.Rational functions can be tricky because they may not be defined for all values of \(x\). This happens because the denominator, \(q(x)\), cannot be zero. In the context of our exercise, when we tried to evaluate \(\left(\frac{g}{f}\right)(0)\), we saw that \(f(0)\) was zero, making the function undefined. Polynomials are easier to deal with, but the division in rational functions requires extra care, especially when determining if a function has any undefined points.

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input value. It's like plugging numbers into a formula to find the outcome. Given a function, you substitute the input value into the equation wherever the variable appears. This tells you what the function yields for that input.In our example, we evaluated two functions: - For \(g(x) = 2x - 1\), when \(x = 0\), the output was \(g(0) = -1\).- For \(f(x) = x^2 + 3x\), when \(x = 0\), the output was \(f(0) = 0\).These evaluations are crucial because they help us determine the result when functions are involved in compositions or other operations. Always ensure to plug in the numbers correctly, as a minor mistake in substitution can lead to incorrect results.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations such as addition, subtraction, multiplication, and division. They are fundamental in mathematics because they express relationships in a concise and systematic way. These expressions sometimes involve powers or roots of variables.Taking our exercise into account, \(f(x) = x^2 + 3x\) and \(g(x) = 2x - 1\) are typical algebraic expressions. Each expression is formed by variables and constants that are combined using arithmetic operations.Understanding these components:

• \(x^2\) is a term with a variable raised to a power.
• \(3x\) is a term consisting of a coefficient (3) multiplied by a variable (x).
• -1 in \(2x - 1\) is a constant term subtracted from the expression.

Recognizing these basic parts within expressions can improve your understanding and manipulation of functions in algebra.