Problem 86
Question
If \(f(x)\) is defined on \((0,1)\), then the domain of definition of \(f\left(e^{x}\right)+f(\log |x|)\) is (A) \((-e,-1)\) (B) \((-e,-1) \cup(1, e)\) (C) \((-\infty,-1)\) (D) \((-e, e)\)
Step-by-Step Solution
Verified Answer
The domain of definition is \((-e, -1)\), so the answer is (A).
1Step 1: Analyze the inner functions
We need to analyze the domain constraints of the functions involved. The function \( f(x) \) is defined on the interval \( (0,1) \). Hence, both \( f(e^x) \) and \( f(\log |x|) \) must satisfy these domain constraints.
2Step 2: Consider \( f(e^x) \)
For \( f(e^x) \) to be defined, \( e^x \) must lie within the domain \( (0,1) \). The inequality \( 0 < e^x < 1 \) implies \( x < 0 \) because \( e^x \) is always positive and less than 1 for negative \( x \). This means \( x \) can be any value from \( (-\infty, 0) \).
3Step 3: Consider \( f(\log |x|) \)
For \( f(\log |x|) \) to be defined, \( \log |x| \) must lie in the interval \( (0,1) \). This means \( 0 < |x| < e \). It implies that \( x \) lies between \( -e \) and \( -1 \) or between \( 1 \) and \( e \), covering intervals \((-e, -1) \cup (1, e)\).
4Step 4: Combine the interval constraints
Finding the intersection of constraints from Steps 2 and 3, we see \( x \) must satisfy: \((-\infty, 0) \cap ((-e, -1) \cup (1, e))\). Thus, \( x \) must belong to \((-e, -1)\).
5Step 5: Select the corresponding answer
Considering that the valid domain for \( x \) is \( (-e, -1) \), the correct choice corresponds to option (A).
Key Concepts
Exponential FunctionsLogarithmic FunctionsInterval Notation
Exponential Functions
Exponential functions are mathematical expressions where the independent variable, typically denoted as \( x \), is located in the exponent. These functions have the general form \( f(x) = a^x \), where \( a \) is a positive constant called the base. In our context, \( a \) is often \( e \), the mathematical constant roughly equal to 2.718. This results in the function \( e^x \).
Features of exponential functions include rapid growth or decay. For example, when \( x \) is positive, \( e^x \) increases rapidly; when \( x \) is negative, \( e^x \) approaches zero without becoming negative. This makes the function always positive regardless of \( x \).
When working with exponential functions like \( f(e^x) \), understanding their range is crucial. Since \( e^x > 0 \) for every real \( x \), we say it's undefined for negative values. In domains, this becomes important as errors arise if calculated boundaries aren't strictly observed.
Features of exponential functions include rapid growth or decay. For example, when \( x \) is positive, \( e^x \) increases rapidly; when \( x \) is negative, \( e^x \) approaches zero without becoming negative. This makes the function always positive regardless of \( x \).
When working with exponential functions like \( f(e^x) \), understanding their range is crucial. Since \( e^x > 0 \) for every real \( x \), we say it's undefined for negative values. In domains, this becomes important as errors arise if calculated boundaries aren't strictly observed.
Logarithmic Functions
Logarithmic functions are the opposites of exponential functions. A logarithm asks the question, "To what power must a given base be raised to produce a specific number?" The common logarithm in natural contexts is the natural logarithm, denoted as \( \log x \), which uses the base \( e \).
The logarithmic function \( y = \log |x| \) implies checking only positive values of \( x \) due to \( |x| \) (absolute value), ensuring it remains non-negative. From our example, \( \log |x| \) takes its value in the interval \( (0, 1) \), meaning that any input \( x \) must result in a log greater than 0 but less than 1.
Characteristic behavior of logarithmic functions include slower growth compared to exponential functions. For domain considerations, \( \log x \) is undefined for non-positive values, asking that any \( x \) results in positive inputs under the log operation.
The logarithmic function \( y = \log |x| \) implies checking only positive values of \( x \) due to \( |x| \) (absolute value), ensuring it remains non-negative. From our example, \( \log |x| \) takes its value in the interval \( (0, 1) \), meaning that any input \( x \) must result in a log greater than 0 but less than 1.
Characteristic behavior of logarithmic functions include slower growth compared to exponential functions. For domain considerations, \( \log x \) is undefined for non-positive values, asking that any \( x \) results in positive inputs under the log operation.
Interval Notation
Interval notation is a concise way of representing subsets of real numbers. It describes the set of numbers between two endpoints. The format \((a, b)\) denotes all numbers greater than \(a\) and less than \(b\), not including \(a\) and \(b\).
When solving problems involving mixed intervals like \\((-e, -1) \cup (1, e)\), understanding their union operation is essential. The union "\( \cup \)" means all numbers in either one of the intervals are included. It combines separate parts of the solution set.
Another crucial concept is intersection. When combining constraints from exponential and logarithmic functions, only values fitting both conditions are selected. In this case, intersecting conditions refine the domain to match real-world constraints needed for correct function operation and problem-solving.
When solving problems involving mixed intervals like \\((-e, -1) \cup (1, e)\), understanding their union operation is essential. The union "\( \cup \)" means all numbers in either one of the intervals are included. It combines separate parts of the solution set.
Another crucial concept is intersection. When combining constraints from exponential and logarithmic functions, only values fitting both conditions are selected. In this case, intersecting conditions refine the domain to match real-world constraints needed for correct function operation and problem-solving.
Other exercises in this chapter
Problem 83
If \(g(x)=1+\sqrt{x}\) and \(f[g(x)]=3+2 \sqrt{x}+x\), then \(f(x)=\) \(\begin{array}{llll}\text { (A) } 1+2 x^{2} & \text { (B) } 2+x^{2} & \text { (C) } 1+x\e
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If \(a\) and \(b\) are natural numbers and \(f(x)=\sin \left(\sqrt{a^{2}-3}\right) x+\cos \left(\sqrt{b^{2}+7}\right) x\) is periodic with finite fundamental pe
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The function \(f(x)=\frac{\sin ^{101} x}{\left[\frac{x}{\pi}\right]+\frac{1}{2}}\), where \([x]\) denotes the integral part of \(x\), is (A) an odd function (B)
View solution