/*! 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 25 Suppose \(r\) is a small positiv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 25

Suppose \(r\) is a small positive number. Estimate the slope of the line containing the points \(\left(e^{2}, 6\right)\) and \(\left(e^{2+r}, 6+r\right)\)

Short Answer

Expert verified
The slope of the line containing the points \((e^2, 6)\) and \((e^{2+r}, 6+r)\) can be estimated using the formula \(m = \frac{y2 - y1}{x2 - x1}\). Substituting the points, we get \(m = \frac{r}{e^{2+r} - e^2}\). Since r is a small positive number, we can assume that \(e^{2+r}\approx e^2\). However, this leads to a denominator very close to zero, making the slope approximation undefined.

Step by step solution

01

Identify the Points

The given points are (e^2, 6) and (e^(2+r), 6+r). We will denote these points as (x1, y1) and (x2, y2) respectively.
02

Apply the Slope Formula

To find the slope of the line, we can use the formula: \[m = \frac{y2 - y1}{x2 - x1}\] By substituting the coordinates of both points into the formula, we can determine the slope of the line.
03

Substitute the Coordinates

Substitute the given points into the slope formula: \[m = \frac{(6+r) - 6}{(e^{2+r}) - e^2}\]
04

Simplify the Expression

Now simplify the expression: \[m = \frac{r}{e^{2+r} - e^2}\]
05

Make an Approximation

Since r is a small positive number, we can assume that e^(2+r) is very close to e^2. So the denominator will become a very small number (almost negligible). Hence we can approximate the slope as: \[m = \frac{r}{e^{2+r} - e^2} \approx \frac{r}{0}\] However, as we cannot divide by zero, the approximation of the slope is undefined.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Estimate the indicated value without using a calculator. $$ e^{0.0013} $$

Suppose \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) are the endpoints of a line segment. (a) Show that the distance between the point \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) and the endpoint \(\left(x_{1}, y_{1}\right)\) equals half the length of the line segment. (b) Show that the distance between the point \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) and the endpoint \(\left(x_{2}, y_{2}\right)\) equals half the length of the line segment.

Explain why the two previous problems imply that \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) is the midpoint of the line segment with endpoints \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\)

Estimate the indicated value without using a calculator. $$ e^{-0.00046} $$

Suppose \(x\) is a positive number. (a) Explain why \(x^{t}=e^{t \ln x}\) for every number \(t\) (b) Explain why $$ \frac{x^{t}-1}{t} \approx \ln x $$ if \(t\) is close to 0 [Part (b) of this problem gives another illustration of why the natural logarithm deserves the title "natural".]
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.