/*! 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 5 Use the linear function \(f(c)+\... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the linear function \(f(c)+\) \(f^{\prime}(c)(x-c)\) to approximate the given quantity. Recall from the previous section that \(\frac{d}{d x}(\sqrt{x})=\frac{1}{2 \sqrt{x}}, \frac{d}{d x}(\ln x)=\frac{1}{x},\) and \(\frac{d}{d x}\left(x^{-1}\right)=-x^{-2} .\) Use your computer or calculator to find the actual answer, and compare your approximation to this. (Of course, the answer given by your computer or calculator is also an approximation.) $$ \ln 0.95 $$

Short Answer

Expert verified
Approximate: \(-0.05\); Actual: \(-0.05129\). Close approximation.

Step by step solution

01

Identify Function and Point for Approximation

We want to approximate \( \ln 0.95 \) using a linear approximation. The linear approximation formula is \( f(c) + f'(c)(x-c) \). Here, \( f(x) = \ln x \). We need to choose a point \( c \) near 0.95 for which we know \( \ln c \) and \( f'(c) \). A convenient choice is \( c = 1 \) since \( \ln 1 = 0 \).
02

Find Derivative at the Chosen Point

Calculate the derivative of the function, \( f'(x) = \frac{1}{x} \). At \( x = 1 \), \( f'(1) = \frac{1}{1} = 1 \).
03

Apply Linear Approximation Formula

Now, use the approximation formula: \( f(c) + f'(c)(x-c) \). Substituting \( c = 1 \), \( f(1) = 0 \), \( f'(1) = 1 \), and \( x = 0.95 \), we get \( \ln 0.95 \approx 0 + 1(0.95 - 1) \). Simplify this to get \( \ln 0.95 \approx -0.05 \).
04

Calculate the Actual Value

Use a calculator to find the actual value of \( \ln 0.95 \). This gives approximately \( -0.05129 \).
05

Compare the Approximation and Actual Value

The estimated value \(-0.05\) is very close to the actual value \(-0.05129\). The linear approximation provides a reasonably accurate estimate.

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Key Concepts

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

Understanding the Derivative
The derivative is a core concept in calculus that describes how a function changes as its input changes. Think of it like the slope of a line on a graph. It tells you the rate at which one quantity changes with respect to another. In this exercise, we used the derivative of the natural logarithm function, which is represented mathematically as \( f'(x) = \frac{1}{x} \).
This derivative tells us how fast the natural logarithm function \( \ln x \) changes as \( x \) changes. By calculating the derivative at a specific point \( c \), we can determine the slope of the tangent line to the curve at that point. This slope is crucial for creating a linear approximation. Remember:
  • The derivative gives you the instant rate of change.
  • It's essential for approximating functions linearly.
  • It helps describe how functions behave locally around a point.
Exploring the Natural Logarithm
The natural logarithm, denoted \( \ln x \), is a special type of logarithm with the base of the number \( e \), where \( e \approx 2.71828 \). It's used in various fields like science, engineering, and mathematics because of its natural growth properties. When we say \( \ln x \), we are asking "what power should \( e \) be raised to in order to get \( x \)?"
In our exercise, we needed to approximate \( \ln 0.95 \). By applying the concept of linear approximation, we were able to get a close estimate using a nearby point where the calculation is straightforward, such as \( c = 1 \). Since \( \ln 1 = 0 \), it's an excellent choice for such approximations. Here are a few facts about the natural logarithm:
  • \( \ln x \) simplifies calculations involving exponential growth.
  • It is the inverse function of the exponential function \( e^x \).
  • The tangent to the curve \( \ln x \) at any point provides insight into its local behavior.
Function Approximation Simplified
Function approximation allows us to estimate complex functions with simpler ones. Linear approximation, in particular, uses the idea of tangent lines to approximate the value of a function near a certain point. We use the formula \( f(c) + f'(c)(x-c) \), where \( c \) is a known value.
This formula gives you a "close enough" estimation of the function value at a nearby point \( x \). It's like creating a straight line that locally mirrors the curve of the function, allowing us to substitute hard calculations with simpler arithmetic. For example, in this exercise, we used the linear approximation to estimate \( \ln 0.95 \), because calculating \( \ln \) can be tricky without a calculator. Here's what makes function approximation useful:
  • Simplifies calculation of complex functions by using basic math.
  • Provides a way to estimate values where precise calculation is difficult.
  • Widely used in scientific calculations to save time and effort.

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Most popular questions from this chapter

Learning The time \(T\) it takes to memorize a list of \(n\) items is given by \(T(n)=3 n \sqrt{n-3}\). Find the approximate change in time required from memorizing a list of 12 items to memorizing a list of 14 items. Find the slope of the tangent line at \(n=12\) by using your computer or graphing calculator to plot the tangent line at the appropriate point, or use a screen with smaller and smaller dimensions until the graph looks like a straight line. (See the Technology Resource Manual.)

Barton and colleagues \(^{61}\) studied how the size of agricultural cooperatives were related to their variable cost ratio (variable cost over assets as a percent). The data is given in the following table. $$ \begin{array}{|l|lllllll|} \hline x & 2.48 & 3.51 & 2.50 & 3.37 & 3.77 & 6.33 & 4.74 \\ \hline y & 11.7 & 16.8 & 25.4 & 16.9 & 17.9 & 22.8 & 16.7 \\ \hline \end{array} $$ Here \(x\) is the assets in millions of dollars, and \(y\) is the variable cost ratio. a. Use quadratic regression to find the best-fitting quadratic function using least squares. b. Graph on your computer or graphing calculator using a screen with dimensions [0,9.4] by \([0,30] .\) Have your grapher find the derivative when \(x\) is \(1.5,3,4.5,6,\) and 6.6. Note the slope, and relate this to the rate of change. On the basis of this model, describe what happens to the rate of change of the variable cost ratio as assets increase.

Find the limits graphically. Then confirm algebraically. $$ \lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2} $$

Elliott \(^{62}\) studied the temperature affects on the alder fly. He collected in his laboratory in 1969 the data shown in the following table relating the temperature in degrees Celsius to the number of pupae successfully completing pupation. $$ \begin{array}{|l|cccccc|} \hline t & 8 & 10 & 12 & 16 & 20 & 22 \\ \hline y & 15 & 29 & 41 & 40 & 31 & 6 \\ \hline \end{array} $$ Here \(t\) is the temperature in degrees Celsius, and \(y\) is the number of pupae successfully completing pupation. a. Use quadratic regression to find \(y\) as a function of \(T\). Graph using a window with dimensions [6,24.8] by [0,60] b. Have your computer or graphing calculator find the numerical derivative when \(x\) is \(10,12,15,18,\) and \(20 .\) Relate this number to the slope of the tangent line to the curve and to the rate of change. Interpret what each of these numbers means. On the basis of this model, describe what happens to the rate of change of number of pupae completing pupation as temperature increases.

Find, without graphing, where each of the given functions is continuous. $$ f(x)=\left\\{\begin{array}{ll} x & \text { if } x<1 \\ 2 & \text { if } x=1 \\ x-1 & \text { if } x>1 \end{array}\right. $$

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