Problem 19
Question
Compute and simplify the difference quotient of the function. $$f(x)=7 x+2$$
Step-by-Step Solution
Verified Answer
Answer: The simplified difference quotient of this function is $$7$$.
1Step 1: 1. Write down the function and difference quotient formula
We have the function $$f(x) = 7x + 2$$, and we need to compute the difference quotient given by the formula:
$$\frac{f(x+h) - f(x)}{h}$$
2Step 2: 2. Evaluate f(x+h) and f(x)
In order to compute the difference quotient, we first need to find $$f(x+h)$$ and $$f(x)$$.
To find $$f(x+h)$$, replace every occurrence of $$x$$ in the function with $$x + h$$:
$$f(x+h) = 7(x+h) + 2 = 7x + 7h + 2$$
Now we have:
$$f(x) = 7x + 2$$ and $$f(x+h) = 7x + 7h + 2$$
3Step 3: 3. Substitute f(x+h) and f(x) into the difference quotient formula
Now we'll substitute our values for $$f(x+h)$$ and $$f(x)$$ into the difference quotient formula:
$$\frac{f(x+h) - f(x)}{h} = \frac{(7x + 7h + 2) - (7x + 2)}{h}$$
4Step 4: 4. Simplify the numerator and denominator
Now, we simplify the numerator by removing the brackets and combining like terms:
$$\frac{(7x + 7h + 2) - (7x + 2)}{h} = \frac{7x + 7h + 2 - 7x - 2}{h} = \frac{7h}{h}$$
5Step 5: 5. Simplify the difference quotient
Finally, we simplify the fraction by canceling the common terms in the numerator and denominator:
$$\frac{7h}{h} = 7$$
We find that the simplified difference quotient of the function $$f(x) = 7x + 2$$ is:
$$7$$.
Key Concepts
Simplify ExpressionsAlgebraic FunctionsPrecalculus
Simplify Expressions
Simplifying expressions is essential for understanding and solving algebraic problems effectively. In the context of the difference quotient, which is a fundamental concept in calculus, simplifying an expression allows you to observe the rate at which a function changes. Let's take the function
When we simplify, we aim to make the expression as straightforward as possible by combining like terms and reducing fractions. For instance, after calculating the difference quotient
From this simplification, the expression
f(x) = 7x + 2 as an example.When we simplify, we aim to make the expression as straightforward as possible by combining like terms and reducing fractions. For instance, after calculating the difference quotient
\(\frac{f(x+h) - f(x)}{h}\), we get to a point where we need to simplify the numerator \(7x + 7h + 2 - 7x - 2\). Notice how the terms 7x and -7x cancel each other out, as do +2 and -2. This leaves us with 7h in the numerator.From this simplification, the expression
\(\frac{7h}{h}\) becomes clear. Since the h terms in the numerator and denominator cancel each other, we are left with the simplified form of 7, demonstrating how simplification is the process of removing complexity to reveal the core value of an expression.Algebraic Functions
Algebraic functions are composed of combinations of polynomial expressions, using arithmetic operations and variables. The difference quotient is a tool used with algebraic functions to analyze the behavior of these functions around a particular input value. Taking the algebraic function
An important skill in dealing with algebraic functions is the ability to manipulate them algebraically. This means substituting in different values for
f(x) = 7x + 2, the core idea is to understand how this function changes as x changes, which is what the difference quotient measures.An important skill in dealing with algebraic functions is the ability to manipulate them algebraically. This means substituting in different values for
x, expanding brackets, and simplifying terms, as seen in the steps of the difference quotient where we substitute x+h for x and simplify. In precalculus and calculus, we explore how these functions graphically represent lines, curves, and more complex shapes, and the difference quotient helps us to grasp their rates of change before we delve into derivatives.Precalculus
Precalculus serves as the foundation for the study of calculus, and it encompasses a wide range of mathematical concepts including algebraic functions, trigonometry, and the theory of equations. Simplifying expressions and understanding the behavior of algebraic functions, as illustrated with the difference quotient, are key precalculus skills.
Students must be comfortable performing operations such as simplification and function evaluation that lead up to the concept of the derivative in calculus. By mastering these precalculus skills, students can more readily understand the concept of the limit and the instantaneous rate of change — both are fundamental to calculus. The difference quotient itself is essentially a prelude to the derivative, offering a numerical expression of the slope of a secant line that approximates the slope of the tangent line as
Students must be comfortable performing operations such as simplification and function evaluation that lead up to the concept of the derivative in calculus. By mastering these precalculus skills, students can more readily understand the concept of the limit and the instantaneous rate of change — both are fundamental to calculus. The difference quotient itself is essentially a prelude to the derivative, offering a numerical expression of the slope of a secant line that approximates the slope of the tangent line as
h approaches zero, which is a cornerstone of differential calculus.Other exercises in this chapter
Problem 18
Determine whether the equation defines \(y\) as a function of \(x\) or defines \(x\) as a function of \(y\) $$x^{2}+2 x y+y^{2}=0$$
View solution Problem 19
Use algebra to find the inverse of the given one-to-one function. $$f(x)=\frac{1}{2 x+1}$$
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Find a single viewing window that shows complete graphs of the functions \(f, g,\) and \(h.\) $$\begin{aligned}&f(x)=.125 x^{3}-.25 x^{2}-1.5 x+5\\\&g(x)=f(x)-5
View solution Problem 19
Find \((g \circ f)(3),(f \circ g)(1),\) and \((f \circ f)(0)\) $$f(x)=x, \quad g(x)=-3$$
View solution