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Chapter 4: Problem 2

For \(f(x)=1 / x\) a. How close must \(x\) be to 0.5 in order that \(f(x)\) is within 0.01 of \(2 ?\) b. How close must \(x\) be to 3 in order to insure that \(\frac{1}{x}\) be within 0.01 of \(\frac{1}{3} ?\) c. How close must \(x\) be to 0.01 in order to insure that \(\frac{1}{x}\) be within 0.1 of \(100 ?\)

Short Answer

Expert verified
a) \(0.4975 < x < 0.5025\); b) \(2.94 < x < 3.13\); c) \(0.0099901 < x < 0.0100101\).

Step by step solution

01

Understand the Problem - Part a

We need to find how close \(x\) should be to 0.5 so that \(f(x) = \frac{1}{x}\) is within 0.01 of 2. This inequality can be written as \( |f(x) - 2| < 0.01 \).
02

Set Up the Inequality - Part a

Rewriting the inequality in terms of \(f(x)\), we have:\[\left| \frac{1}{x} - 2 \right| < 0.01\] This inequality can be split into two parts: \(\frac{1}{x} - 2 < 0.01\) and \(\frac{1}{x} - 2 > -0.01\).
03

Solve the First Inequality - Part a

For \(\frac{1}{x} - 2 < 0.01\), simplify to find:\[ \frac{1}{x} < 2.01\]Invert both sides to solve for \(x\) (remember to flip the inequality):\[ x > \frac{1}{2.01}\]
04

Solve the Second Inequality - Part a

For \(\frac{1}{x} - 2 > -0.01\), simplify to find:\[\frac{1}{x} > 1.99\]Invert both sides to solve for \(x\) (flip the inequality again):\[x < \frac{1}{1.99}\]
05

Determine the Range for x - Part a

Combine the inequalities from Steps 3 and 4 to find the range for \(x\):\[\frac{1}{2.01} < x < \frac{1}{1.99}\]Calculating these gives:\[0.4975 < x < 0.5025\]
06

Understand the Problem - Part b

For part b, find the range for \(x\) such that \(\left| \frac{1}{x} - \frac{1}{3} \right| < 0.01\).
07

Set Up the Inequality - Part b

Rewriting the inequality, we have:\[\left| \frac{1}{x} - \frac{1}{3} \right| < 0.01\]Which implies two parts: \(\frac{1}{x} - \frac{1}{3} < 0.01\) and \(\frac{1}{x} - \frac{1}{3} > -0.01\).
08

Solve the First Inequality - Part b

Simplifying \(\frac{1}{x} - \frac{1}{3} < 0.01\) leads to:\[\frac{1}{x} < \frac{0.34}{0.33} \approx 0.34\]Invert the inequality:\[x>\frac{1}{0.34}\]
09

Solve the Second Inequality - Part b

Simplifying \(\frac{1}{x} - \frac{1}{3} > -0.01\) leads to:\[\frac{1}{x} > \frac{0.32}{0.33} \approx 0.32\]Invert the inequality:\[x<\frac{1}{0.32}\]
10

Determine the Range for x - Part b

Combine the inequalities from Steps 8 and 9 to find the range for \(x\):\[\frac{1}{0.34} < x < \frac{1}{0.32}\]Calculating these gives:\[2.94 < x < 3.13\]
11

Understand the Problem - Part c

Find how close \(x\) needs to be to 0.01 so that \(\frac{1}{x}\) is within 0.1 of 100. This means:\[\left| \frac{1}{x} - 100 \right| < 0.1\]
12

Set Up the Inequality - Part c

Rewriting the inequality gives:\[\frac{1}{x} - 100 < 0.1\] and \[\frac{1}{x} - 100 > -0.1\].
13

Solve the First Inequality - Part c

For \(\frac{1}{x} - 100 < 0.1\):\[\frac{1}{x} < 100.1\]Invert the inequality:\[x > \frac{1}{100.1}\]
14

Solve the Second Inequality - Part c

For \(\frac{1}{x} - 100 > -0.1\):\[\frac{1}{x} > 99.9\]Invert the inequality:\[x < \frac{1}{99.9}\]
15

Determine the Range for x - Part c

Combine the inequalities from Steps 13 and 14 to find the range for \(x\):\[\frac{1}{100.1} < x < \frac{1}{99.9}\]Calculating these gives:\[0.0099901 < x < 0.0100101\]
16

Conclusion

In each part, we've found the range of \(x\) which ensures \(f(x) = \frac{1}{x}\) is within given bounds of the target value.

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Key Concepts

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

Inequalities
Inequalities help us understand the relationship between numbers, showing when one number is larger or smaller than another. In calculus, inequalities are crucial for defining ranges and conditions under which certain propositions hold true. Let's delve into understanding inequalities within this context.
  • Absolute Inequalities: These involve expressions like \( |f(x) - a| < ext{some value} \), signifying that the difference between function \( f(x) \) and a number \( a \) must remain within a certain distance. It forms the basis for ensuring functions are close to a desired value.
  • Solving Inequalities: This involves dividing the inequality into parts: \( f(x) - a < ext{value} \) and \( f(x) - a > - ext{value} \). Solving each gives possible values for \( x \) that satisfy the condition.
  • Combining Inequalities: After solving separately, combine the findings to establish a range for \( x \). For example, determining \( 0.4975 < x < 0.5025 \) for a function \( f(x) = \frac{1}{x} \).
Understanding these concepts in calculus allows you to assess how functions behave within specific bounds.
Inverse Functions
An inverse function reverses the operations of the original function. If a function \( f \) maps a number \( x \) to \( y \), then its inverse \( f^{-1} \) maps \( y \) back to \( x \). In the exercise, the function \( f(x) = \frac{1}{x} \) is considered.
  • Identifying Inverse Functions: To find an inverse, solve \( y = f(x) \) for \( x \). For \( f(x) = \frac{1}{x} \), rearranging gives \( x = \frac{1}{y} \), making its inverse \( f^{-1}(y) = \frac{1}{y} \).
  • Verifying Inverses: Apply the original and inverse function successively as \( f(f^{-1}(y)) = x \) and \( f^{-1}(f(x)) = x \). This checks the reversal accurately returns the initial value.
  • Application in Problems: In dealing with inverse functions, such as here with \( f(x) = \frac{1}{x} \), the examination is vital for understanding how function constraints are interpreted in terms of its inverse.
Grasping inverse functions is critical for exploring how outputs of functions revert to inputs under inverse operations.
Limits
Limits in calculus describe the behavior of a function as it approaches a particular value or infinity. They help establish the function鈥檚 tendency, showing what value it gets close to under specific conditions. Let's explore the notion of limits.
  • Concept of Limits: Informally, limits determine how \( f(x) \) behaves as \( x \) approaches some number. Mathematically expressed as \( \lim_{x \to a} f(x) \).
  • Approaching Values: For \( f(x) = \frac{1}{x} \), understand how it behaves as \( x \) approaches a number like 0. This means observing values \( f(x) \) nears, even if \( x \) isn鈥檛 actually reaching \( a \).
  • Significance in Calculus Problems: In the exercise, determine ranges such as how \( x \) nears numbers like 0.5 or 3 while maintaining \( f(x) \) within bounds like 0.01 to ensure understanding of function behavior.
Comprehending limits ensures insights into the function's approach toward particular points or infinities.
Rational Functions
Rational functions consist of ratios of polynomials expressed as \( \frac{P(x)}{Q(x)} \). These are a fundamental part of calculus when identifying behaviors, especially near undefined points or when making approximations. We'll understand these through function \( f(x) = \frac{1}{x} \).
  • Characteristics of Rational Functions: These include analysis of domains and asymptotes. The function is undefined where \( Q(x) = 0 \), which for \( f(x) = \frac{1}{x} \) is at \( x = 0 \), showcasing vertical asymptotes.
  • Behavioral Analysis: Look at approaches near undefined points and horizontal asymptotic behaviors as \( x \) tends to infinity. For \( \frac{1}{x} \), it veers toward zero as \( x \) grows large.
  • Role in Problem Solving: The exercise's use of \( f(x) = \frac{1}{x} \) explores how close \( x \) must be to certain values to confine \( f(x) \) within specific bounds. This informs about potential function discontinuity or behavior asymptotes.
Understanding rational functions in calculus challenges students to predict the functions' traits and boundary interactions effectively.

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Most popular questions from this chapter

Exercise 4.1 .5 a. Draw the graph of \(y_{1}\). b. Find a number \(A\) such that the graph of \(y_{2}\) is continuous. $$ \text { a. } \quad y_{1}(x)=\left\\{\begin{array}{rll} x^{2} & \text { for } x<2 \\ 3-x & \text { for } 2 \leq x \end{array} \quad\right. \text { b. } y_{2}(x)=\left\\{\begin{aligned} x^{2} & \text { for } x<2 \\ A-x & \text { for } 2 \leq x \end{aligned}\right. $$

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Exercise \(4.1 .9 \quad\) a. Draw the graph of a function, \(f,\) defined on the interval [1,3] such that \(f(1)=-2\) and \(f(3)=4\) b. Does your graph intersect the X-axis? c. Draw a graph of of a function, \(f,\) defined on the interval [1,3] such that \(f(1)=-2\) and \(f(3)=4\) that does not intersect the X-axis. Be sure that its X-projection is all of [1,3] . d. Write equations to define a function, \(f,\) on the interval [1,3] such that \(f(1)=-2\) and \(f(3)=4\) and the graph of \(f\) does not intersect the X-axis. e. There is a theorem that asserts that the function you just defined must be discontinuous at some number in [1,3] . Identify such a number for your example. The preceding exercise illustrates a general property of continuous functions called the intermediate value property. Briefly it says that a continuous function defined on an interval that has both positive and negative values on the interval, must also be zero somewhere on the interval. In language of graphs, the graph of a continuous function defined on an interval that has a point below the X-axis and a point above the X-axis must intersect the X-axis. The proof of this property requires more than the familiar properties of addition, multiplication, and order of the real numbers. It requires the completion property of the real numbers, Axiom \(5.2 .1^{2}\).
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