Problem 8
Question
(a) find the simplified form of the difference quotient and then (b) complete the following table. $$ \begin{array}{|c|l|l|} \hline x & h & \frac{f(x+h)-f(x)}{h} \\ \hline 5 & 2 & \\ \hline 5 & 1 & \\ \hline 5 & 0.1 & \\ \hline 5 & 0.01 & \\ \hline \end{array} $$ $$ f(x)=\frac{2}{x} $$
Step-by-Step Solution
Verified Answer
The simplified difference quotient is \( \frac{-2}{x(x+h)} \). For (x, h) = (5, 2), (5, 1), (5, 0.1), (5, 0.01): outputs are \(-\frac{2}{35}\), \(-\frac{1}{15}\), \(-0.078431\), \(-0.07984\) respectively.
1Step 1: Identify the Function and Form of the Difference Quotient
The function given is \( f(x) = \frac{2}{x} \). The difference quotient for a function \( f \) is \( \frac{f(x+h) - f(x)}{h} \). Our goal is to simplify this expression for the given function.
2Step 2: Substitute the Function into the Difference Quotient
Substitute \( f(x) = \frac{2}{x} \) into the difference quotient: \[ \frac{ \frac{2}{x+h} - \frac{2}{x} }{h} \]. This involves finding a common denominator for the fractions in the numerator.
3Step 3: Simplify the Numerator
The common denominator for \( \frac{2}{x+h} \) and \( \frac{2}{x} \) is \( x(x+h) \). Express each fraction with this common denominator: \( \frac{2x - 2(x+h)}{x(x+h)} \). Simplify the numerator: \( 2x - 2x - 2h = -2h \).
4Step 4: Complete the Simplification
The expression now is \( \frac{-2h}{h(x+h)} \). Cancel out the \( h \) from the numerator and denominator, assuming \( h eq 0 \): \( \frac{-2}{x(x+h)} \). This is the simplified form of the difference quotient.
5Step 5: Calculate Each Value in the Table
For each entry in the table, substitute \( x=5 \) and the respective \( h \) values into \( \frac{-2}{x(x+h)} \):* For \( h = 2 \): \( \frac{-2}{5(5+2)} = \frac{-2}{35} \).* For \( h = 1 \): \( \frac{-2}{5(6)} = \frac{-2}{30} = \frac{-1}{15} \).* For \( h = 0.1 \): \( \frac{-2}{5(5.1)} \approx -0.078431 \).* For \( h = 0.01 \): \( \frac{-2}{5(5.01)} \approx -0.07984 \).
Key Concepts
Difference QuotientLimit ProcessRational FunctionsAlgebraic Simplification
Difference Quotient
The difference quotient is a fundamental concept in calculus. It measures the average rate of change of a function, which is represented as \( \frac{f(x+h) - f(x)}{h} \). This formula essentially calculates the slope of the secant line that passes through the points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of the function.
When working with the difference quotient, the primary goal is to simplify the expression by substituting the given function. For example, if you have a function like \( f(x) = \frac{2}{x} \), you'd substitute this into the difference quotient to find \( \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \).
Understanding and simplifying this expression lays the groundwork for exploring the behavior of the function as the input changes. It's a foundational step towards grasping the concept of derivatives.
When working with the difference quotient, the primary goal is to simplify the expression by substituting the given function. For example, if you have a function like \( f(x) = \frac{2}{x} \), you'd substitute this into the difference quotient to find \( \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \).
Understanding and simplifying this expression lays the groundwork for exploring the behavior of the function as the input changes. It's a foundational step towards grasping the concept of derivatives.
Limit Process
The limit process is an essential part of calculus that helps us understand what happens to a function as its variable approaches a particular value. It allows us to find derivatives, which represent the rate of change or the slope of a tangent line to the curve.
In the context of the difference quotient, the limit process comes into play as we investigate what occurs when \( h \), the difference in the input, approaches zero. By doing this, we are effectively finding the instantaneous rate of change, denoted by the derivative.
Using our earlier simplified form \( \frac{-2}{x(x+h)} \), as \( h \) approaches zero, we get closer to understanding how the function behaves at a specific point. This is crucial for deeper mathematical analysis and real-world applications, such as calculating speed or optimizing profits.
In the context of the difference quotient, the limit process comes into play as we investigate what occurs when \( h \), the difference in the input, approaches zero. By doing this, we are effectively finding the instantaneous rate of change, denoted by the derivative.
Using our earlier simplified form \( \frac{-2}{x(x+h)} \), as \( h \) approaches zero, we get closer to understanding how the function behaves at a specific point. This is crucial for deeper mathematical analysis and real-world applications, such as calculating speed or optimizing profits.
Rational Functions
Rational functions, like \( f(x) = \frac{2}{x} \), are quotients of polynomial functions. They are an important class of functions in calculus because they can exhibit interesting behavior, such as asymptotes and discontinuities.
When dealing with rational functions, you often work with algebraic expressions that involve division, which can become very complex. Simplifying these expressions is crucial for applying calculus concepts like the difference quotient.
Rational functions tend to have unique characteristics. For example, they can approach infinity as \( x \) approaches zero if the denominator goes to zero, showcasing the importance of understanding both the function values and limits.
When dealing with rational functions, you often work with algebraic expressions that involve division, which can become very complex. Simplifying these expressions is crucial for applying calculus concepts like the difference quotient.
Rational functions tend to have unique characteristics. For example, they can approach infinity as \( x \) approaches zero if the denominator goes to zero, showcasing the importance of understanding both the function values and limits.
Algebraic Simplification
Algebraic simplification is a critical skill that involves manipulating expressions to make them easier to work with. In calculus, this means breaking down complex expressions to understand their behavior and limit properties.
For instance, in simplifying the difference quotient \( \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \), finding a common denominator helps combine fractions to form a single expression. This leads to simplifying the numerator into something like \( -2h \), which can then cancel with the \( h \) in the denominator, reducing to a more manageable form like \( \frac{-2}{x(x+h)} \).
Mastering these simplifications allows you to explore calculus concepts more deeply and solve a wider range of problems with ease.
For instance, in simplifying the difference quotient \( \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \), finding a common denominator helps combine fractions to form a single expression. This leads to simplifying the numerator into something like \( -2h \), which can then cancel with the \( h \) in the denominator, reducing to a more manageable form like \( \frac{-2}{x(x+h)} \).
Mastering these simplifications allows you to explore calculus concepts more deeply and solve a wider range of problems with ease.
Other exercises in this chapter
Problem 7
Classify each statement as either true or false. If \(\lim _{x \rightarrow 4} F(x)\) exists, then \(F\) must be continuous at \(x=4\)
View solution Problem 7
Complete each of the following statements. The notation__________ is read “the limit, as x approaches 2 from the right.”
View solution Problem 8
Find \(\frac{d y}{d x}\). $$ y=2 x^{15} $$
View solution Problem 8
Find \(d^{2} y / d x^{2}\) $$ y=6 x-3 $$
View solution