Problem 32
Question
Assume \(h \neq 0 .\) Compute and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h} $$ $$f(x)=1 / x$$
Step-by-Step Solution
Verified Answer
Answer: The simplified difference quotient for the function \(f(x) = 1/x\) is \(\frac{-1}{x(x+h)}\).
1Step 1: Plug in the function
Plug in \(f(x)=1/x\) to the difference quotient formula:
$$
\frac{1/(x+h) - 1/x}{h}
$$
2Step 2: Simplify the numerator
Find a common denominator and combine the fractions in the numerator:
$$
\frac{\frac{x - (x + h)}{x(x + h)}}{h}
$$
3Step 3: Simplify the expression further
Now, we simplify the expression in the numerator and the denominator by cancelling out the variable \(h\):
$$
\frac{\frac{-h}{x(x + h)}}{h}
$$
4Step 4: Final simplification
Finally, we cancel out the \(h\) terms to get the simplified difference quotient:
$$
\frac{-1}{x(x+h)}
$$
Key Concepts
CalculusFunction AnalysisRational Functions
Calculus
Calculus is a branch of mathematics that studies how things change. One of the core tools in calculus is the concept of the difference quotient. It helps us understand the rate at which a function changes at any given point. When we talk about the difference quotient, we are often looking at how a small change in the input of a function (often denoted as \( h \)) affects the output of the function. This measures the average rate of change of the function over the interval \([x, x + h]\).
In this exercise, we're calculating the difference quotient for the function \( f(x) = \frac{1}{x} \). This involves taking the expression \( \frac{f(x+h) - f(x)}{h} \) and simplifying it. This process is important because it lays the groundwork for taking a derivative. The derivative of a function is an advanced concept that tells us exactly how a function's output changes with a small change in input, and it is derived from the limit of this difference quotient as \( h \) approaches zero.
In this exercise, we're calculating the difference quotient for the function \( f(x) = \frac{1}{x} \). This involves taking the expression \( \frac{f(x+h) - f(x)}{h} \) and simplifying it. This process is important because it lays the groundwork for taking a derivative. The derivative of a function is an advanced concept that tells us exactly how a function's output changes with a small change in input, and it is derived from the limit of this difference quotient as \( h \) approaches zero.
Function Analysis
Function analysis involves understanding the behavior of a function. This means analyzing its examples, patterns, continuity, and limits, among other properties. With the rational function \( f(x) = \frac{1}{x} \), analyzing its behavior includes understanding its domain and range, how it behaves as \( x \) approaches certain values, and how it can be manipulated algebraically.
In our exercise, we focused on the simplification process. Firstly, this involved substituting into the difference quotient. Then, it requires manipulating the fractions to find a common denominator and eventually canceling the variable \( h \). These steps are crucial for students to practice as they reveal how functions behave algebraically and help develop a deeper understanding of their structure, which is essential when dealing with more complex calculus problems later.
In our exercise, we focused on the simplification process. Firstly, this involved substituting into the difference quotient. Then, it requires manipulating the fractions to find a common denominator and eventually canceling the variable \( h \). These steps are crucial for students to practice as they reveal how functions behave algebraically and help develop a deeper understanding of their structure, which is essential when dealing with more complex calculus problems later.
- Identifying the domain of a function
- Finding common denominators
- Simplifying expressions
Rational Functions
Rational functions are ratios of polynomial expressions. They can have asymptotes, which occur where the function approaches infinity. This happens in the function \( f(x) = \frac{1}{x} \) as well, particularly around \( x = 0 \), where the function is undefined.
Understanding rational functions is key to simplifying complex expressions such as those found in the difference quotient. In the exercise, a rational function was simplified by combining fractions with a common denominator, and by canceling common terms. These are basic techniques in simplifying rational expressions, but they are foundational for dealing with more challenging calculus problems that involve rational functions in their contexts.
Understanding rational functions is key to simplifying complex expressions such as those found in the difference quotient. In the exercise, a rational function was simplified by combining fractions with a common denominator, and by canceling common terms. These are basic techniques in simplifying rational expressions, but they are foundational for dealing with more challenging calculus problems that involve rational functions in their contexts.
- Cancelling terms when possible
- Understanding when and how rational expressions are undefined
- Analyzing asymptotic behavior
Other exercises in this chapter
Problem 31
Assume \(h \neq 0 .\) Compute and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h} $$ $$f(x)=\sqrt{x}$$
View solution Problem 32
Verify that \((f \circ g)(x)=x\) and \((g \circ f)(x)=x\) for every \(x\) $$f(x)=2 x^{3}-5, \quad g(x)=\sqrt[3]{\frac{x+5}{2}}$$
View solution Problem 32
Find an equation that expresses the area \(A\) of a circle as a function of its (a) radius \(r\) (b) diameter \(d\)
View solution Problem 33
Find the approximate intervals on which the function is increasing, those on which it is decreasing, and those on which it is constant. $$f(x)=-x^{3}-8 x^{2}+8
View solution