Ch. 1 - Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo

Briggs 3rd Edition

Ch. 1 - Functions

Problema 1.1.68

Capitolo 1, Problema 1.1.68

Simplify the difference quotient ƒ(x+h)-ƒ(x)/h
ƒ(x) = (x)/(x+1)

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First, understand the difference quotient: it is a formula used to approximate the derivative of a function. The general form is (f(x+h) - f(x)) / h.

Identify the function f(x) given in the problem, which is f(x) = x / (x + 1).

Calculate f(x + h) by substituting x + h into the function: f(x + h) = (x + h) / (x + h + 1).

Substitute f(x + h) and f(x) into the difference quotient formula: ((x + h) / (x + h + 1) - x / (x + 1)) / h.

Simplify the expression by finding a common denominator for the fractions in the numerator, then simplify the entire expression step by step, focusing on algebraic manipulation to reduce the complexity.

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Difference Quotient

The difference quotient is a formula used to calculate the average rate of change of a function over an interval. It is expressed as (ƒ(x+h) - ƒ(x))/h, where h represents a small change in x. This concept is fundamental in calculus as it leads to the definition of the derivative, which measures the instantaneous rate of change.

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The Quotient Rule

Function Evaluation

Function evaluation involves substituting a specific value into a function to determine its output. In this case, we need to evaluate the function ƒ(x) = x/(x+1) at both x and x+h. Understanding how to correctly substitute and simplify expressions is crucial for manipulating the difference quotient.

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Evaluating Composed Functions

Simplification Techniques

Simplification techniques in algebra involve reducing expressions to their simplest form, often by factoring, combining like terms, or canceling common factors. In the context of the difference quotient, applying these techniques will help in simplifying the expression after substituting ƒ(x+h) and ƒ(x), making it easier to analyze the limit as h approaches zero.

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Initial Value Problems

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