/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Make a table of values for the f... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 38

Make a table of values for the function \(F(x)=(x+2) /(x-2)\) at the points \(x=1.2, x=11 / 10, x=101 / 100, x=1001 / 1000\) \(x=10001 / 10000,\) and \(x=1 .\) a. Find the average rate of change of \(F(x)\) over the intervals \([1, x]\) for each \(x \neq 1\) in your table. b. Extending the table if necessary, try to determine the rate of change of \(F(x)\) at \(x=1\) .

Short Answer

Expert verified
The rate of change of \(F(x)\) at \(x=1\) is approximately \(-4\).

Step by step solution

01

Calculate F(x) for given points

Substitute each value of x into the function \(F(x) = \frac{x+2}{x-2}\) and calculate the result.- For \(x = 1.2\): \[F(1.2) = \frac{(1.2)+2}{(1.2)-2} = \frac{3.2}{-0.8} = -4\]- For \(x = \frac{11}{10}\): \[F\left(\frac{11}{10}\right) = \frac{\frac{11}{10} + 2}{\frac{11}{10} - 2} = \frac{\frac{31}{10}}{-\frac{9}{10}} = -\frac{31}{9}\]- For \(x = \frac{101}{100}\): \[F\left(\frac{101}{100}\right) = \frac{\frac{101}{100} + 2}{\frac{101}{100}-2} = \frac{\frac{301}{100}}{-\frac{99}{100}} = -\frac{301}{99}\]- For \(x = \frac{1001}{1000}\): \[F\left(\frac{1001}{1000}\right) = \frac{\frac{1001}{1000} + 2}{\frac{1001}{1000} - 2} = \frac{\frac{3001}{1000}}{-\frac{999}{1000}} = -\frac{3001}{999}\]- For \(x = \frac{10001}{10000}\): \[F\left(\frac{10001}{10000}\right) = \frac{\frac{10001}{10000} + 2}{\frac{10001}{10000} - 2} = \frac{\frac{30001}{10000}}{-\frac{9999}{10000}} = -\frac{30001}{9999}\]
02

Find average rate of change for each interval [1, x]

The average rate of change of \(F(x)\) over the interval \([1, x]\) is given by the formula \( \frac{F(x) - F(1)}{x - 1} \), where \( F(1) = -3 \) based on the limit evaluation.- For \(x = 1.2\): \[\frac{-4 - (-3)}{1.2 - 1} = \frac{-1}{0.2} = -5\]- For \(x = \frac{11}{10}\): \[\frac{-\frac{31}{9} - (-3)}{\frac{1}{10}} = \frac{-\frac{4}{9}}{\frac{1}{10}} = -\frac{40}{9}\]- For \(x = \frac{101}{100}\): \[\frac{-\frac{301}{99} - (-3)}{\frac{1}{100}} = \frac{-\frac{4}{99}}{\frac{1}{100}} = -\frac{400}{99}\]- For \(x = \frac{1001}{1000}\): \[\frac{-\frac{3001}{999} - (-3)}{\frac{1}{1000}} = \frac{-\frac{4}{999}}{\frac{1}{1000}} = -\frac{4000}{999}\]- For \(x = \frac{10001}{10000}\): \[\frac{-\frac{30001}{9999} - (-3)}{\frac{1}{10000}} = \frac{-\frac{4}{9999}}{\frac{1}{10000}} = -\frac{40000}{9999}\]
03

Determine the rate of change at x = 1

Notice that as \(x\) gets closer to 1, the average rates of change approach a consistent pattern. Calculate the limit of the average rate of change as \(x \to 1\):Considering the values:- -5 for \(x = 1.2\)- Close values like \(-\frac{40}{9} \approx -4.44\), increasing negatively as denominators get smaller.Guessing from the pattern, the average rate of change becomes more negative and approaches \(-4\) as \(x\) approaches 1 from the right. For precise calculation, observe that derivatives will confirm the instantaneous rate of change is \(-4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a process of finding the value of a given function for a particular input. In this exercise, we evaluated the function \( F(x) = \frac{x+2}{x-2} \) at specific points such as \( x = 1.2, \frac{11}{10}, \frac{101}{100} \), and more. To find \( F(x) \) at each point, substitute each \( x \) value into the function and perform the arithmetic operations. This way, you will get a set of outputs, or values of the function, corresponding to each input. Function evaluation helps understand the behavior of the function at different points, which is important for understanding changes over intervals. Remember, when substituting values into a function, pay attention to the arithmetic operations to avoid small mistakes which can lead to incorrect results.
Derivatives
Derivatives provide a tool to determine the rate at which function values change. If you think about driving a car, the derivative answers the question: "How fast is the car going at an exact moment?" In calculus, the derivative of a function at a point gives the slope of the tangent line to the graph of the function at that point. It's a fundamental concept used to find how things change instantly. In this particular exercise, by examining the consistent pattern of the average rate of change as values of \( x \) approach 1, we can deduce the derivative. Although we didn't calculate it directly, we determined the instantaneous rate of change at \( x = 1 \) using the concept of limits.
Limits
Limits help us understand what happens to a function when the input approaches a particular value. They are essential when calculating derivatives and determining the behavior of functions at points where standard evaluation might not work directly. In this exercise, by evaluating the function as \( x \) gets closer to 1, we estimate its behavior and rate of change without directly substituting \( x = 1 \) due to complications like division by zero. As we calculate average rates of change for points getting increasingly close to 1, the limit assists to show the trend, approaching a consistent outcome. This concept is crucial as it allows us to predict what would occur even when direct calculations are complicated or impossible.
Instantaneous Rate of Change
The instantaneous rate of change is akin to capturing the exact speed at a moment when you're moving. In mathematical terms, it's the value of the derivative at a specific point on a curve. It's like taking a snapshot of a moving object's speedometer. In this scenario, as we examined successive average rates of change and their limit as \( x \) approached 1, we inferred that the instantaneous rate of change at \( x = 1 \) approaches \(-4\). It is important to compute this rate because it reflects how the function changes narrowly around a specific point, unlike average rate of change which considers broader intervals. This gives us a precise understanding of the function's behavior at an exact point rather than over an interval.

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Most popular questions from this chapter

In Exercises \(11-18,\) find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$ f(x)=x^{2}+1, \quad(2,5) $$

If the product function \(h(x)=f(x) \cdot g(x)\) is continuous at \(x=0\) , must \(f(x)\) and \(g(x)\) be continuous at \(x=0 ?\) Give reasons for your answer.

Find the limits in Exercises \(29-34 .\) Are the functions continuous at the point being approached? $$ \lim _{t \rightarrow 0} \cos \left(\frac{\pi}{\sqrt{19-3 \sec 2 t}}\right) $$

Find the limits in Exercises \(29-34 .\) Are the functions continuous at the point being approached? $$ \lim _{y \rightarrow 1} \sec \left(y \sec ^{2} y-\tan ^{2} y-1\right) $$

In Exercises \(39-42,\) sketch the graph of a function \(y=f(x)\) that satisfies the given conditions. No formulas are required- just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.) $$ \begin{array}{l}{f(0)=0, f(1)=2, f(-1)=-2, \lim _{x \rightarrow-\infty} f(x)=-1, \text { and }} \\ {\lim _{x \rightarrow \infty} f(x)=1}\end{array} $$
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