Question

Given \(f(x)=\frac{1}{x}\) and \(\left(\frac{f}{g}\right)(x)=\frac{x+1}{x^{2}-x},\) find the function \(g\)

Short answer

The function \(g(x)\) is \( \frac{x-1}{x+1} \).

Step-by-step solution

Understand the Given Functions

We are given two functions: 1. \(f(x)=\frac{1}{x}\) 2. \(\left( \frac{f}{g} \right) (x) = \frac{x+1}{x^{2}-x}\). We need to find the function \(g(x)\).

Set Up the Equation

The expression \(\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)}=\frac{x+1}{x^{2}-x}\). Substitute \(f(x)=\frac{1}{x}\): \[ \frac{\frac{1}{x}}{g(x)} = \frac{x+1}{x^{2}-x} \]

Solve for g(x)

Rearrange the equation to isolate \(g(x)\): \[ g(x) = \frac{\frac{1}{x}}{\frac{x+1}{x^{2}-x}} \] Simplify the expression: \[ g(x) = \frac{\frac{1}{x}}{\frac{x+1}{x(x-1)}} = \frac{1/x}{(x+1)/(x(x-1))} \] Multiplying by the reciprocal: \[ g(x) = \frac{1/x}{(x+1)/(x(x-1))} = \frac{1/x} \cdot \frac{x(x-1)}{x+1} = \frac{x(x-1)}{x(x+1)} = \frac{x-1}{x+1} \]

Key concepts

rational functions

Rational functions are a type of function that is the ratio of two polynomials. This means that a rational function looks like a fraction where both the numerator and the denominator are polynomials. For example, the function given in the exercise, \(\frac{1}{x}\) and \(\frac{x+1}{x^2-x}\), are both rational functions.
The general form of a rational function is \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials. It's important to note that the denominator, \(Q(x)\), cannot be zero because division by zero is undefined.
Rational functions can have various behaviors, such as vertical asymptotes, horizontal asymptotes, or holes. These behaviors arise where the denominator equals zero or where the function is not defined. Understanding these properties helps us analyze and graph rational functions.

function operations

Function operations refer to arithmetic operations that can be performed on functions, such as addition, subtraction, multiplication, division, and composition. In the exercise, we specifically deal with division of functions. For function division, if we have two functions \(f(x)\) and \(g(x)\), the division of these functions is written as \(\frac{f}{g}(x)\) and represents the expression \(\frac{f(x)}{g(x)}\).
It's crucial to ensure that the denominator function \(g(x)\) is not zero because it would make the division undefined. To perform division of functions:

• Write the expression as a fraction.
• Simplify if possible.

In our exercise, we initially have \(\frac{1/x}{(x+1)/(x^2-x)}\). Such complex fractions can be simplified by multiplying by the reciprocal of the denominator. This step often makes it easier to identify and cancel out common factors.

simplifying expressions

Simplifying expressions involves reducing them to their simplest form. This usually means removing any unnecessary terms, combining like terms, or factoring polynomials when possible. In our exercise, simplifying the given expression was a key step to find the function \(g(x)\).
Here’s the simplified process: Initially, we had the expression \(g(x) = \frac{\frac{1}{x}}{\frac{x+1}{x(x-1)}}\). By multiplying the numerator by the reciprocal of the denominator, we turned the complex fraction into a simpler one:

• Convert the denominator fractions into a single fraction, \( \frac{x+1}{x(x-1)} \).
• Multiply by the reciprocal: \ g(x) = \frac{1/x}{(x+1)/(x(x-1))} = \frac{1/x} \times \frac{x(x-1)}{x+1} \.
• Simplify: \ g(x) = \frac{x(x-1)}{x(x+1)} = \frac{x-1}{x+1} \.

Simplifying the expressions not only makes them easier to work with, but it also allows you to accurately identify the resulting function, which is necessary for solving the problem correctly.