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Chapter 2: Problem 20

Calculate how close \(x\) needs to be to \(c\) to force \(f(x)\) to be within 0.01 of \(\ell\) $$ f(x)=2 x-3, c=-2, \ell=-7 $$

Short Answer

Expert verified
x must be within 0.005 units of c = -2.

Step by step solution

01

Recognize the Problem

To find how close the value of \( x \) needs to be to \( c \) so that \( f(x) \) stays within 0.01 units of \( \ell \), we need to set up an equation using the definition of limits. This involves making \( |f(x) - \ell| < 0.01 \).
02

Substitute Values into the Inequality

The inequality to work with is \( |f(x) - \ell| < 0.01 \). Substitute \( f(x) = 2x - 3 \) and \( \ell = -7 \) to get: \(|2x - 3 + 7| < 0.01\). Simplifying this gives \( |2x + 4| < 0.01 \).
03

Solve the Inequality for x

Now, solve the simplified inequality \( |2x + 4| < 0.01 \):1. Remove the absolute value by splitting it into two inequalities: - \( 2x + 4 < 0.01 \) - \( 2x + 4 > -0.01 \).2. Solve each inequality separately: - For \( 2x + 4 < 0.01 \): \( 2x < 0.01 - 4 \) \( 2x < -3.99 \) \( x < -1.995 \) - For \( 2x + 4 > -0.01 \): \( 2x > -0.01 - 4 \) \( 2x > -4.01 \) \( x > -2.005 \).
04

Combine the Solutions for x

The solutions from Step 3 give the interval for \( x \):\(-2.005 < x < -1.995\). This means \( x \) must fall within this interval to make \( f(x) \) close enough to \( \ell \).
05

Relate x to c

Now that we have the interval \(-2.005 < x < -1.995\), we need to express this in terms of \( c = -2 \):- Calculate the distance from \( c \) to each bound of the interval: - \( |-2.005 - (-2)| = 0.005 \) - \( |-1.995 - (-2)| = 0.005 \).Thus, \( x \) needs to be within 0.005 units of \( c = -2 \) to ensure \( f(x) \) stays within 0.01 of \( \ell = -7 \).

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Key Concepts

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

Inequalities in Calculus
When studying limits in calculus, inequalities play a crucial role. Inequalities help us define the bounds within which a function must lie to meet specific conditions. In the given exercise, we need to find the range of values for \( x \) that will keep \( f(x) \) within a certain range of \( \ell \). This is done using the inequality \( |f(x) - \ell| < 0.01 \).

To solve this, we first substitute the values of \( f(x) = 2x - 3 \) and \( \ell = -7 \) into the inequality. This transforms our expression to \( |2x + 4| < 0.01 \).

The absolute value sign leads us to set up two separate inequalities:
  • \( 2x + 4 < 0.01 \)
  • \( 2x + 4 > -0.01 \)
Solving each of these, we realize that \( x \) must satisfy both conditions to meet the limit requirement. This highlights how inequalities are used to carve out acceptable intervals for function behavior.
Absolute Value in Calculus
Absolute value is a fundamental concept in calculus because it measures the size of a number regardless of its sign. In our exercise, the expression \( |f(x) - \ell| < 0.01 \) uses absolute value to ensure \( f(x) \) stays within a close distance to \( \ell \).

By rewriting \( f(x) - \ell \) into the inequality \( |2x + 4| < 0.01 \), we ask how close the expression \( 2x + 4 \) must be to zero. This is pivotal because it treats the situation symmetrically 鈥� without any preference for either side of zero 鈥� allowing us to solve for two scenarios:
  • \( 2x + 4 < 0.01 \)
  • \( 2x + 4 > -0.01 \)
In the process, absolute value ensures we're considering \( f(x) \)'s proximity to \( \ell \) equally on both the higher and lower end. This symmetric nature is what makes absolute values a reliable tool in defining limits and bounds in calculus.
Distance and Interval in Calculus
Understanding distance and interval concepts in calculus is crucial when dealing with limits. The main task is determining how close \( x \) needs to be to \( c \) to satisfy our function's limit condition.

In this exercise, solving the inequalities leads us to discover that \( x \) must lie between \(-2.005\) and \(-1.995\). This defines a very specific interval where the limit condition holds true. The interval \(-2.005 < x < -1.995\) indicates the precise range within which \( x \) must stay.

To relate this to the given point \( c = -2 \), we calculate the distance from \( c \) to the interval bounds. Both distances, \(|-2.005 - (-2)|\) and \(|-1.995 - (-2)|\), equal 0.005.
  • This means \( x \) must be within 0.005 units from \( c \) which satisfies our initial condition on \( f(x) \).
This detailed understanding of distance and intervals is essential in determining "how close" a value of \( x \) needs to be in limit problems.

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Most popular questions from this chapter

Plot \(y=\exp (x)\) for \(0 \leq x \leq 2\). Let \(P(c)\) denote the point \((c, \exp (c))\) on the graph. The purpose of this exercise is to graphically explore the relationship between \(\exp (c)\) and the slope of the tangent line at \(P(c) .\) For \(c=1 / 2,1\), and \(3 / 2,\) calculate the slope \(m(c)\) of the secant line that passes through the pair of points \(P(c-0.001)\) and \(P(c+0.001) .\) For each \(c,\) calculate \(|\exp (c)-m(c)|\) to see that \(m(c)\) is a good approximation of \(\exp (c) .\) Add the three secant lines to your viewing window. For each of \(c=1 / 2,1,\) and \(3 / 2,\) add to the viewing window the line through \(P(c)\) with slope \(\exp (c)\). As we will see in Chapter \(3,\) these are the tangent lines at \(P(1 / 2), P(1)\) and \(P(3 / 2)\). It is likely that they cannot be distinguished from the secant lines in your plot.

An assertion is made about a function \(f\) that is defined on a closed, bounded interval. If the statement is true, explain why. Otherwise, sketch a function \(f\) that shows it is false. (Note: \(|f|\) is defined by \(|f|(x)=|f(x)| .)\) If \(f\) is continuous, then \(|f|\) is continuous.

An assertion is made about a function \(f\) that is defined on a closed, bounded interval. If the statement is true, explain why. Otherwise, sketch a function \(f\) that shows it is false. (Note: \(|f|\) is defined by \(|f|(x)=|f(x)| .)\) If \(f^{2}\) is continuous, then \(f\) is continuous.

Locate, to four decimal places of accuracy, the maximum and minimum values of the function \(h(x)=\) \(x^{4}-5 x^{3}+7 x+9\) on the interval [0,4]

A particle moves on an axis. Its position \(p(t)\) at time \(t\) is given. For a positive \(h,\) the average velocity over the time interval \([2,2+h]\) is \(\bar{v}(h)=\frac{p(2+h)-p(2)}{h}\) a. Numerically determine \(v_{0}=\lim _{h \rightarrow 0+} \bar{v}(h)\). b. How small does \(h\) need to be for \(\bar{v}(h)\) to be between \(v_{0}\) and \(v_{0}+0.1 ?\) c. How small does \(h\) need to be for \(\bar{v}(h)\) to be between \(v_{0}\) and \(v_{0}+0.01 ?\) $$ p(t)=9 t^{2} /(t+1) $$
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