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Chapter 3: Problem 61

Use a graphing utility to graph the function. $$f(x)=|\ln x|$$

Short Answer

Expert verified
The graph of the function \(f(x)=|\ln x|\) resembles a 'V', starting at point (1,0). The right arm of the 'V' represents the graph of \(y = \ln x\) for \(x > 1\), while the left arm represents the reflection of the graph of \(y = - \ln x\) for \(0 < x < 1\).

Step by step solution

01

Understand the Basic Logarithmic Function

Firstly, it's necessary to understand how the basic natural logarithm function, \( \ln x \), behaves. The function is undefined for \( x \leq 0 \), and it crosses the x-axis at \( x=1 \). This function increases as x increases, but at a decreasing rate.
02

Understand the Absolute Value Function

The absolute value function, denoted by \( |x| \), makes all negative values turn positive, but leaves positive values unchanged. Essentially, it reflects any point below the x-axis to the corresponding point above.
03

Combine the Functions

Now we can combine our understanding of these two functions to graph \( f(x) = |\ln x| \). For \( x > 1 \), the function behaves as a normal logarithm function, since \( \ln x \) is positive for these values. However, for \( 0 < x < 1 \), where \( \ln x \) is negative, the absolute value will reflect these points above the x-axis.
04

Graph the Function Using a Graphing Tool

Finally, graph the function using a graphing utility. You should see a curve increasing from (1,0), extending rightwards, and a reflection of this curve below the x-axis. The function is undefined at \( x = 0 \) and for negative x values.

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Key Concepts

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

Natural logarithm
The natural logarithm, denoted as \( \ln x \), is a fundamental concept in mathematics used to describe logarithms with base \( e \). Here, \( e \) is an irrational constant approximately equal to \( 2.71828 \). The function \( \ln x \) is only defined for positive values of \( x \). When \( x = 1 \), the value of \( \ln x \) is \( 0 \). This function grows as \( x \) increases, but it does so at a decelerating rate. In essence, for every unit increase in \( x \), the increment in \( \ln x \) is smaller. This behavior is significant when graphing because it details how the curve increases from \( (1,0) \).
  • Undefined for \( x \leq 0 \)
  • Increases at a decreasing rate
  • Interesting point: the curve passes through \((1,0)\)
Absolute value
The absolute value function is a straightforward yet powerful tool in mathematics, denoted as \( |x| \). It ensures that all input values through \( x \) are turned to non-negative amounts. For positive \( x \), it remains the same, while any negative \( x \) is reflected over the x-axis, turning positive.
  • Keeps all positive values unchanged
  • Converts all negative values to positive, reflecting them over the x-axis
When combined with functions like \( \ln x \), it creates a symmetrical effect, useful for visualizing the behavior of functions around challenging points, such as when \( x \) is just above 0.
Graphing utilities
Graphing utilities are vital tools in modern mathematics, providing a practical way to visually analyze functions. These can be software tools like graphing calculators or online platforms. They allow students and educators to plot complex functions accurately, saving time and improving comprehension. To graph \( f(x) = |\ln x| \), input this expression into your graphing utility. Here's what the tool can do:
  • Plot nails the curve shape and symmetry
  • Show critical points and their positions
  • Offer insights into function behavior at points like \( (1,0) \)
Using graphing utilities effectively highlights nuances in functions, helping to see reflections or asymmetrical behaviors that are less obvious in manual plots.
Reflections in graphs
Reflections in graphs occur when components of a function mirror across an axis. In \( f(x) = |\ln x| \), this is achieved by the absolute value affecting \( \ln x \). For \( x > 1 \), \( \ln x \) remains positive and the curve follows its typical ascending path. But when \( 0 < x < 1 \), \( \ln x \) would normally dip below the x-axis. The absolute value flips it positively, reflecting across the x-axis to create a symmetric pattern.
  • Mirrors negative portions of a function to positive
  • Produces symmetry across the x-axis
Understanding reflections is crucial in graphing, revealing how functions adapt to different transformations inherent in their expressions.

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Most popular questions from this chapter

Involve the factorial function \(x !\), which is defined for whole numbers \(x\) as $$ x !=\left\\{\begin{array}{ll} 1, & \text { if } x=0 \\ x \cdot(x-1) \cdot(x-2) \cdot \cdots \cdot \cdot 3 \cdot 2 \cdot 1, & \text { if } x \geq 1 \end{array}\right. $$ For example, \(3 !=3 \cdot 2 \cdot 1=6\) and \(5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120\) During the 30 -minute period before a Broadway play begins, the members of the audience arrive at the theater at the average rate of 12 people per minute. The probability that \(x\) people will arrive during a particular minute is given by \(P(x)=\frac{12^{x} e^{-12}}{x !} .\) Find the probability, to the nearest \(0.1 \%\) that a. 9 people will arrive during a given minute. b. 18 people will arrive during a given minute.

The monthly income \(I\), in dollars, from a new product is given by $$ I(t)=24,000-22,000 e^{-0.005 t} $$ where \(t\) is the time, in months, since the product was first put on the market. a. What was the monthly income after the 10 th month and after the 100 th month? b. What will the monthly income from the product approach as the time increases without bound?

The distance \(s\) (in feet) that the object in Exercise 31 will fall in \(t\) seconds is given by \(s=32 t+32\left(e^{-t}-1\right)\) a. Use a graphing utility to graph this equation for \(t \geq 0\) b. Determine, to the nearest 0.1 second, the time it takes the object to fall 50 feet. c. Calculate the slope of the secant line through \((1, s(1))\) and \((2, s(2))\) d. Write a sentence that explains the meaning of the slope of the secant line you calculated in \(c .\)

The number of bass in a lake is given by $$ P(t)=\frac{3600}{1+7 e^{-0.05 t}} $$ -where \(t\) is the number of months that have passed since the lake was stocked with bass. a. How many bass were in the lake immediately after it was stocked? b. How many bass were in the lake 1 year after the lake was stocked? c. What will happen to the bass population as \(t\) increases without bound?

Make use of the factorial function, which is defined as follows. For whole numbers \(n\), the number \(n !\) (which is read "n factorial") is given by $$n !=\left\\{\begin{array}{ll} n(n-1)(n-2) \cdots 1, & \text { if } n \geq 1 \\\1, & \text { if } n=0 \end{array}\right.$$Thus, \(0 !=1\) and \(4 !=4 \cdot 3 \cdot 2 \cdot 1=24\) STIRLING'S FORMULA Stirling's Formula (after James Stir\(\operatorname{lin} g, 1692-1770)\) 1$$n !=\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}$$ is often used to approximate very large factorials. Use Stirling's Formula to approximate \(10 !\), and then compute the ratio of Stirling's approximation of \(10 !\) divided by the actual value of \(10 !\), which is 3,628,800
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