Problem 13
Question
(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=x^{2}-4 x $$
Step-by-Step Solution
Verified Answer
The simplified difference quotient is \( 2x + h - 4 \). Table values: 8, 7, 6.1, and 6.01.
1Step 1: Find the difference quotient
The difference quotient is given by \( \frac{f(x+h)-f(x)}{h} \). First, find \( f(x+h) \). Substitute \( x+h \) into the given function \( f(x) = x^2 - 4x \), resulting in \( f(x+h) = (x+h)^2 - 4(x+h) \). Simplifying, we get \( x^2 + 2xh + h^2 - 4x - 4h \).
2Step 2: Simplify the expression
Now, calculate \( f(x+h) - f(x) \) using \( f(x) = x^2 - 4x \). This gives: \( (x^2 + 2xh + h^2 - 4x - 4h) - (x^2 - 4x) = 2xh + h^2 - 4h \).
3Step 3: Calculate the simplified difference quotient expression
Simplify the expression \( \frac{2xh + h^2 - 4h}{h} \) by dividing each term by \( h \), resulting in \( 2x + h - 4 \). This is the simplified form of the difference quotient.
4Step 4: Compute the difference quotient for each row in the table
For each value of \( h \) in the table, substitute \( x = 5 \) and calculate the difference quotient \( 2x + h - 4 \):- For \( h = 2 \): \( 2(5) + 2 - 4 = 10 + 2 - 4 = 8 \)- For \( h = 1 \): \( 2(5) + 1 - 4 = 10 + 1 - 4 = 7 \)- For \( h = 0.1 \): \( 2(5) + 0.1 - 4 = 10 + 0.1 - 4 = 6.1 \)- For \( h = 0.01 \): \( 2(5) + 0.01 - 4 = 10 + 0.01 - 4 = 6.01 \).
Key Concepts
CalculusFunctionsLimits
Calculus
Calculus is a branch of mathematics that studies how things change. It's all about rates of change and accumulation. At its core, calculus helps us figure out how to deal with continuous change. This is unlike algebra, which typically handles things that are static. One of the fundamental ideas is the concept of a derivative, which measures how a function changes as its input changes.
To understand derivatives, we use the difference quotient. This concept represents the average rate of change of a function over a small interval. Calculus provides the tools to analyze behaviors of complex functions. In our example, the difference quotient is used to find how the function \(f(x) = x^2 - 4x\) changes around a certain point. This is just the beginning as calculus extends to more complex ideas, such as integration and multivariable calculus.
To understand derivatives, we use the difference quotient. This concept represents the average rate of change of a function over a small interval. Calculus provides the tools to analyze behaviors of complex functions. In our example, the difference quotient is used to find how the function \(f(x) = x^2 - 4x\) changes around a certain point. This is just the beginning as calculus extends to more complex ideas, such as integration and multivariable calculus.
Functions
Functions are mathematical constructs that relate an input to an output. In simple terms, a function takes an input, applies some operation to that input, and then gives an output. Functions can be represented in various ways including formulas, graphs, or verbal descriptions.
In the given exercise, we are working with the function \(f(x) = x^2 - 4x\). This function takes an input \(x\) and transforms it using the prescribed operations. One useful feature of functions is their ability to show relationships and help visualize concepts. For instance, in the context of difference quotients, how \(f(x)\) changes as \(x\) changes by a small amount \(h\) is pivotal. Understanding functions allows us to predict outcomes and analyze how different parameters affect the behavior of our mathematical models.
In the given exercise, we are working with the function \(f(x) = x^2 - 4x\). This function takes an input \(x\) and transforms it using the prescribed operations. One useful feature of functions is their ability to show relationships and help visualize concepts. For instance, in the context of difference quotients, how \(f(x)\) changes as \(x\) changes by a small amount \(h\) is pivotal. Understanding functions allows us to predict outcomes and analyze how different parameters affect the behavior of our mathematical models.
Limits
The concept of limits is a cornerstone of calculus and crucial for defining derivatives and integrals. A limit helps us understand the behavior of functions as they approach a particular point. When we talk about limits, what we're usually interested in is what happens to a function’s value as the input gets very close to a certain number.
With the difference quotient \(\frac{f(x+h)-f(x)}{h}\), the limit comes into play as \(h\) approaches zero. This is because we want to know the instantaneous rate of change of the function at a particular point. In the exercise, we see that as \(h\) becomes smaller, the difference quotient gives a better approximation of the derivative at \(x = 5\). The closer \(h\) is to zero, the more precise our understanding of the function's rate of change at that point.
With the difference quotient \(\frac{f(x+h)-f(x)}{h}\), the limit comes into play as \(h\) approaches zero. This is because we want to know the instantaneous rate of change of the function at a particular point. In the exercise, we see that as \(h\) becomes smaller, the difference quotient gives a better approximation of the derivative at \(x = 5\). The closer \(h\) is to zero, the more precise our understanding of the function's rate of change at that point.
Other exercises in this chapter
Problem 12
Consider the function \(f\) given \(b y\) $$ f(x)=\left\\{\begin{array}{ll} x-2, & \text { for } x \leq 3, \\ x-1, & \text { for } x>3. \end{array}\right. $$ If
View solution Problem 12
Use the Theorem on Limits of Rational Functions to find each limit. When necessary, state that the limit does not exist. $$ \lim _{x \rightarrow-2}\left(x^{2}+3
View solution Problem 13
Find \(\frac{d y}{d x}\). $$ y=x^{4}-7 x $$
View solution Problem 13
Find \(f^{\prime \prime}(x)\) $$ f(x)=x^{3}-\frac{5}{x} $$
View solution