/*! 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 42 a. Graph the five functions \(\l... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 42

a. Graph the five functions \(\ln x-2, \ln x-1, \ln x\) \(\ln x+1,\) and \(\ln x+2\) on the window [0,4] by [-3,3] b. Find the slope (using NDERIV or \(d y / d x\) ) of several of the curves at a particular \(x\) -value and check that in each case the slope is the reciprocal of the \(x\) -value. This suggests that the derivative of each function is \(1 / x\) c. Based on part (b), conjecture what is the indefinite integral of the function \(1 / x\) (for \(x>0\) ).

Short Answer

Expert verified
The indefinite integral of \(\frac{1}{x}\) for \(x>0\) is \(\ln |x| + C\).

Step by step solution

01

Understanding the Functions

The functions given are transformations of the natural logarithm function. These transformations are shifts vertically upward or downward from the base function \( \ln x \). Specifically, the functions are: \( \ln x - 2 \), \( \ln x - 1 \), \( \ln x \), \( \ln x + 1 \), and \( \ln x + 2 \).
02

Graphing the Functions

Graph each of the functions: \( \ln x - 2 \), \( \ln x - 1 \), \( \ln x \), \( \ln x + 1 \), and \( \ln x + 2 \) on the window [0,4] by [-3,3]. Each graph will have the same shape, but they will be vertically shifted. For instance, \( \ln x - 2 \) will be shifted 2 units down compared to \( \ln x \), and \( \ln x + 2 \) will be 2 units up.
03

Calculating Slopes Using Derivatives

To find the slope of the curves at a particular \( x \)-value, use the derivative \( \frac{d}{dx}(\ln x) = \frac{1}{x} \). Evaluate this derivative at various \( x \)-values within the domain of the plots. For example, at \( x = 2 \), \( \frac{d}{dx}(\ln x) = \frac{1}{2} \), at \( x = 1 \), \( \frac{d}{dx}(\ln x) = 1 \).
04

Verifying the Relationship

Upon calculation, we notice that the slope \( \frac{1}{x} \) is indeed the reciprocal of the \( x \)-value chosen. This verifies that the derivative of the function \( \ln x \) retains this property across different vertical shifts: \( \ln x - 2, \ln x - 1, \ln x, \ln x + 1, \ln x + 2 \).
05

Conjecturing the Indefinite Integral

Based on the properties observed in part (b), the indefinite integral of the function \( \frac{1}{x} \) for \( x>0 \) is conjectured to be \( \ln |x| + C \), where \( C \) is the constant of integration.

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Key Concepts

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

Derivative
In calculus, a derivative represents how a function changes as its input changes. When you find the derivative, you're essentially finding the rate of change or the slope of the function at any given point. The derivative of a function is a core concept, because it gives insight into how the function behaves. For the natural logarithm function, which is expressed as \( \ln x \), the derivative is \( \frac{1}{x} \).
  • This means at any point \( x \) on the function \( \ln x \), the slope of the tangent line is \( \frac{1}{x} \).
  • Understanding derivatives helps in predicting how functions increase or decrease. For \( x > 0 \), as \( x \) increases, \( \frac{1}{x} \) decreases, showing a slower rate of increase.
  • This derivative is independent of vertical shifts, so whether it's \( \ln x - 2 \) or \( \ln x + 2 \), the slope at any given \( x \) depends solely on \( \frac{1}{x} \).
Understanding this concept is crucial for graph analysis and applied calculus where rate of change is key.
Graphing Functions
Graphing functions helps us visualize how a mathematical expression behaves across different values. When we graph functions like \( \ln x - 2 \), \( \ln x - 1 \), \( \ln x \), \( \ln x + 1 \), and \( \ln x + 2 \), we notice these are simply vertical shifts of the \( \ln x \) function.
  • Each function retains the shape of the basic \( \ln x \) curve.
  • The graph shifts upwards if a positive constant is added, and shifts downwards with a negative constant.
  • Visualizing these shifts can help in better understanding how the function transforms without affecting its core properties like the rate of change (which is determined by its derivative).
Graphing is an essential tool in calculus to predict and confirm observations about functions and their behavior over an interval.
Indefinite Integral
The indefinite integral of a function is essentially the reverse operation of taking a derivative; it finds an antiderivative. For the function \( \frac{1}{x} \), its indefinite integral gives the original function before differentiation (within a constant factor).
  • The indefinite integral of \( \frac{1}{x} \) for \( x > 0 \) is \( \ln |x| + C \), where \( C \) is the constant of integration.
  • This shows that taking the integral of \( \frac{1}{x} \) brings us back to the logarithmic framework, indicating a deep relationship between exponentials and logarithms.
  • Having the constant \( C \) is crucial as it accounts for any vertical shift in the graph of the original function.
Understanding indefinite integrals is important in calculus as it allows us to determine the original function from its rate of change.

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Most popular questions from this chapter

Suppose that a company found its rate of revenue (dollars per day) and its (lower) rate of costs (also in dollars per day). If you integrated "upper minus lower" over a month, describe the meaning of the number that you would find.

After \(t\) hours of work, a medical technician can carry out T-cell counts at the rate of \(2 t^{2} e^{-t / 4}\) tests per hour. How many tests will the technician process during the first eight hours (time 0 to time 8 )?

A company's marginal revenue function is \(M R(x)=700 x^{-1}\) and its marginal cost function is \(M C(x)=500 x^{-1}\) (both in thousands of dollars), where \(x\) is the number of units \((x>1)\). Find the total profit from \(x=200\) to \(x=300\)

Choose the correct answer. \(\int x^{-1} d x=?\) a. \(\ln |x|+C\) b. \(\frac{1}{0} x^{0}+C\) c. \(x^{-1} x+C\)

\(85-94 .\) The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int(x+1)(x-5)^{4} d x $$
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