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. \(f(x)=-3 x+1\)

Short answer

The difference quotient is \( -3 \).

Step-by-step solution

- Define the function

The given function is defined as: \[ f(x) = -3x + 1 \]

- Apply the function to \( x + h \)

To find the difference quotient, first substitute \( x + h \) into the function: \[ f(x + h) = -3(x + h) + 1 \] Simplifying this gives: \[ f(x + h) = -3x - 3h + 1 \]

- Compute the numerator

The numerator of the difference quotient is \( f(x + h) - f(x) \). Substitute the definitions for \( f(x + h) \) and \( f(x) \): \[ f(x + h) - f(x) = (-3x - 3h + 1) - (-3x + 1) \] Simplify the expression: \[ f(x + h) - f(x) = -3x - 3h + 1 + 3x - 1 \] \[ f(x + h) - f(x) = -3h \]

- Form and simplify the difference quotient

Now divide the numerator by \( h \): \[ \frac{f(x + h) - f(x)}{h} = \frac{-3h}{h} \] Simplify the fraction: \[ \frac{f(x + h) - f(x)}{h} = -3 \]

Key concepts

Understanding Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols.
In algebra, symbols (like x or y) are used to represent numbers in equations and functions.
This helps to generalize arithmetic operations.

In our difference quotient problem, we are working with a linear function, which is an equation of the form:



In this example, we have the function:










LtAlgebra teaches us about different properties of numbers and their relationships, and helps simplify expressions and solve equations.

Exploring Functions

A function is a relationship between two sets of values, where each input (from one set) is related to exactly one output (from another set).
Functions can be represented in various ways, such as equations, graphs, or tables.
In a mathematical notation, if we have a function \(f\), we write \(f(x)\), where \(x\) is the input. The output is determined by the function rule.

For instance, in our exercise, the function \(f\) is given by:




To understand this function better, let's break it down: 'x' represents the input value.
'-3' is the coefficient that multiples 'x'.
'+1' is a constant term added to the product.

By substituting different values for 'x', we can determine the function output.
For example, if \(x = 2\), then \(f(2) = -3(2) + 1 = -6 + 1 = -5\).
Understanding this linear function helps us in simplifying the difference quotient by accurately substituting and replacing values.

Grasping Limits

Limits are a foundational concept in calculus that describe the behavior of a function as the input approaches a particular value.
When working with the difference quotient, we are interested in how the function behaves as 'h' approaches 0.

In mathematical notation, we represent a limit as:


In our step-by-step solution, after simplifying the difference quotient, we observed:

This indicates that no matter how small we make 'h', the result will always approach -3.
The concept of limits is crucial because it helps in understanding the instantaneous rate of change of functions, leading to derivatives in calculus.
Knowing limits, we can formally define the derivative of a function, which is the limit of the difference quotient as 'h' approaches 0.


The limit concept not only assists in simplifying the mathematical process but also in grasping the behavior and changes of functions at precise points.