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Chapter 4: Problem 200

For the following exercises, state the domain and the vertical asymptote of the function. $$g(x)=\ln (3-x)$$

Short Answer

Expert verified
Domain: \(x < 3\); Vertical Asymptote: \(x = 3\).

Step by step solution

01

Understanding the Domain

The domain of a function describes all the possible values of x for which the function is defined. Since we are dealing with the natural logarithm, the expression inside the logarithm must be greater than zero, i.e., \(3-x > 0\).
02

Solving for Domain

To find the domain, solve the inequality \(3-x > 0\). Rearranging gives \(x < 3\). Thus, the domain of the function \(g(x) = \ln(3-x)\) is all x such that \(x < 3\). In interval notation, this is \((-fty, 3)\).
03

Identifying the Vertical Asymptote

A vertical asymptote typically occurs where the function is undefined due to the input approaching a critical value. Here, \(x = 3\) is where \(3-x = 0\) (thus, \(g(x)\) is undefined for \(x \geq 3\)). Therefore, the vertical asymptote is \(x = 3\).
04

Conclusion

The domain of \(g(x) = \ln(3-x)\) is \(x < 3\) and the vertical asymptote is \(x = 3\).

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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 is a fundamental mathematical function denoted by \( \ln(x) \). It is the inverse of the exponential function \( e^x \), which means that \( \ln(e^x) = x \). The natural logarithm has several important properties:
  • Defined for Positive Values: The natural logarithm is only defined for positive values of \( x \). This implies that when dealing with expressions within a logarithm, ensure the expression is greater than zero.

  • Relationship with Exponentials: Since it is the inverse of the exponential, \( \ln(e) = 1 \) and \( \ln(1) = 0 \).

  • Simplifying Expressions: Inequalities involving the logarithm often guide us in determining valid inputs, helping to define function domains.

Understanding these properties is crucial when working through problems involving the natural logarithm, such as identifying the domain of a logarithmic function.
Vertical Asymptote
A vertical asymptote of a function is a vertical line \( x = a \) where the function approaches infinity or negative infinity as \( x \) approaches \( a \). This usually occurs when a function is undefined for a particular value of \( x \).To find a vertical asymptote:
  • Identify where the function is undefined. Often this involves setting the denominator to zero or evaluating critical points like in the expression provided by a logarithm.

  • In the problem involving \( g(x) = \ln(3-x) \), the vertical asymptote is at \( x = 3 \). This is because substituting \( x = 3 \) into the function gives \( \ln(0) \), which is undefined.

  • Approaching the asymptote, the function鈥檚 value increases or decreases without bound, graphically showing the asymptotic behavior near \( x = 3 \).

Recognizing where these vertical asymptotes occur is integral to understanding the behavior of logarithmic and other rational functions.
Inequalities
Inequalities are statements that compare two expressions using symbols such as \( < \), \( > \), \( \leq \), or \( \geq \). Solving inequalities involves finding the range of values that satisfy the given condition.In the context of determining the domain of \( g(x) = \ln(3-x) \):
  • We need to solve the inequality \( 3-x > 0 \) to ensure the logarithm is well-defined. Rearranging gives \( x < 3 \), constraining \( x \) to values less than 3.

  • The solution to the inequality provides all the acceptable values for \( x \), forming the domain of the function.

  • Representing the domain using interval notation, we find it to be \( (-\infty, 3) \), clearly excluding the point where the function becomes undefined (\( x = 3 \)).

Solving inequalities is a practical skill in calculus and algebra, crucial for defining domains and understanding function behaviors.

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Most popular questions from this chapter

For the following exercises, refer to Table 4.28. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {5.1} & {6.3} & {7.3} & {7.7} & {8.1} & {8.6} \\\ \hline\end{array}$$ Graph the logarithmic equation on the scatter diagram.

For the following exercises, use properties of logarithms to evaluate without using a calculator. $$ 2 \log _{9}(3)-4 \log _{9}(3)+\log _{9}\left(\frac{1}{729}\right) $$

For the following exercises, refer to Table 4.30. $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} & {10} \\ \hline f(x) & {8.7} & {12.3} & {15.4} & {18.5} & {20.7} & {22.5} & {23.3} & {24} & {24.6} & {24.8} \\\ \hline\end{array}$$ To the nearest whole number, what is the predicted carrying capacity of the model?

For the following exercises, refer to Table 4.26. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {1125} & {1495} & {2310} & {3294} & {4650} & {6361} \\\ \hline\end{array}$$ Use a graphing calculator to create a scatter diagram of the data.

Use logarithms to solve. \(8 e^{-5 x-2}-4=-90\)
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