Question

Find the difference quotient of \(f\); that is, find \(\frac{f(x+h)-f(x)}{h}, h \neq 0,\) for each function. Be sure to simplify. \(f(x)=\frac{1}{x^{2}+1}\)

Short answer

\(\frac{-(2x+h)}{((x+h)^2 + 1)(x^2 + 1)}\)

Step-by-step solution

Write the Difference Quotient Formula

The difference quotient formula is \ \[\frac{f(x+h) - f(x)}{h}\] \ where \(heq 0\).

Determine \(f(x+h)\)

Substitute \(x+h\) into the function \(f(x)\) to get \(f(x+h)\): \ \[f(x+h) = \frac{1}{(x+h)^2 + 1}\]

Substitute \(f(x)\) and \(f(x+h)\) into the Difference Quotient

Substitute \(f(x) = \frac{1}{x^2 + 1}\) and \(f(x+h) = \frac{1}{(x+h)^2 + 1}\) into the difference quotient formula: \ \[\frac{\frac{1}{(x+h)^2 + 1} - \frac{1}{x^2 + 1}}{h}\]

Combine the Fractions in the Numerator

Combine the fractions \(\frac{1}{(x+h)^2 + 1} - \frac{1}{x^2 + 1}\) using a common denominator: \ \[ \frac{(x^2+1) - ((x+h)^2 + 1)}{((x+h)^2 + 1)(x^2 + 1)} \]

Simplify the Expresson inside the Numerator

Simplify the expression inside the numerator: \ \[ (x^2+1) - ((x+h)^2 + 1) = x^2 + 1 - (x^2 + 2xh + h^2 + 1) = -2xh - h^2 \] \ Therefore, the combined fraction becomes: \ \[ \frac{-2xh - h^2}{((x+h)^2 + 1)(x^2 + 1)} \]

Factor out \(-h\) in the Numerator

Factor out \(-h\) from the numerator: \ \[ \frac{-h(2x+h)}{((x+h)^2 + 1)(x^2 + 1)} \]

Simplify by Cancelling \(h\)

Simplify the fraction by cancelling \(h\) in the numerator and denominator: \ \[ \frac{-h(2x+h)}{h((x+h)^2 + 1)(x^2 + 1)} = \frac{-(2x+h)}{((x+h)^2 + 1)(x^2 + 1)} \]

Key concepts

Functions

In mathematics, a function is a relationship between two sets. Every element in the first set (known as the domain) corresponds to exactly one element in the second set (known as the range). We denote functions by letters such as f, g, and h. For example, in the given problem, we have a function f(x) = \(\frac{1}{x^2 + 1}\). This indicates that for any input x, the output is \(\frac{1}{x^2 + 1}\). Understanding functions helps us better analyze and process relationships between various quantities.

Rational Functions

Rational functions are a specific type of function where the output is the ratio of two polynomials. For example, f(x) = \(\frac{1}{x^2 + 1}\) is a rational function because the numerator is a polynomial (specifically, a constant 1), and the denominator is a polynomial \(x^2 + 1\). Rational functions often have interesting properties and behaviors, particularly around the values that make the denominator zero (the function is not defined at these points). These functions are common in calculus and algebra due to their significance in modeling various real-world scenarios.

Simplification

Simplification is the process of reducing expressions to their simplest form. Simplifying helps in understanding and solving problems more efficiently. When working with the difference quotient, simplification steps are crucial. For example, combining fractions and finding common denominators as seen in the numerator \(\frac{1}{(x + h)^2 + 1} - \frac{1}{x^2 + 1}\). Ensuring each step is simplified correctly makes following the solution process clearer and helps avoid mistakes. Remember that simplifying involves basic algebra, such as factoring and combining like terms.

Variables

Variables are symbols that represent numbers or values in mathematical expressions or equations. In our problem, 'x' and 'h' are variables. 'x' usually denotes the input to the function, while 'h' represents a small change applied to 'x', which is important in calculus when calculating limits, slopes of tangent lines (derivatives), or in this case, the difference quotient. Treating variables correctly is essential. Each substitution, combination, or simplification step must consistently acknowledge these variables' roles to ensure accurate results. Understanding how to manipulate variables and expressions they are part of is a foundational skill in mathematics.