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Find the domain of each function. \((a) f(x)=\frac{1-e^{x^{2}}}{1-e^{1-x^{2}}} \quad (b) f(x)=\frac{1+x}{e^{\cos x}}\)

Short Answer

Expert verified
(a) All real numbers except \( x = 1 \) and \( x = -1 \); (b) all real numbers.

Step by step solution

01

Understanding the Function

We are given the function \( f(x) = \frac{1-e^{x^{2}}}{1-e^{1-x^{2}}} \). To determine its domain, we need to identify values of \( x \) that could make the denominator zero because division by zero is undefined.
02

Finding the Domain of Part (a)

For the given function \( f(x) = \frac{1-e^{x^{2}}}{1-e^{1-x^{2}}} \), the denominator cannot equal zero. Thus, set \( 1-e^{1-x^{2}} = 0 \), which simplifies to \( e^{1-x^{2}} = 1 \). Since the exponential function equals 1 when its exponent is zero, we solve \( 1-x^{2} = 0 \). This gives \( x^{2} = 1 \), leading to \( x = \pm 1 \). The domain of \( f(x) \) is all real numbers except \( x = 1 \) and \( x = -1 \).
03

Understanding the Second Function

For part (b), we are given \( f(x) = \frac{1+x}{e^{\cos x}} \). The denominator should not be zero, but since the exponential function \( e^{\cos x} \) is never zero, this part is always defined.
04

Determining Domain of Part (b)

Since \( e^{\cos x} eq 0 \) for any real number \( x \), the function \( f(x) = \frac{1+x}{e^{\cos x}} \) is defined for all real numbers \( x \). Thus, the domain is all real numbers.

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Key Concepts

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

Exponential function
Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. A classic example is the function \( e^x \), where \( e \) (approximately 2.718) is the base of natural logarithms. Exponential functions are crucial in describing growth and decay processes in various fields such as biology, finance, and physics. They exhibit rapid growth or decay depending on the sign of the exponent.
Commonly, exponential functions can be expressed in the form \( f(x) = a \cdot e^{b \cdot x} \), where \( a \) and \( b \) are constants. When examining exponential functions within a problem, it is essential to ensure that operations applied do not alter their inherent properties. For instance, \( e^{x^2} \) in the problem exercise means an exponentiated parabolic growth, as the exponent is quadratic.
  • They always pass through the point \( (0, a) \) if \( f(x) = a \cdot e^{b \cdot x} \).
  • Their domain is typically all real numbers, but conditions in scenarios (like solving for zero in denominators) can restrict this.
Real numbers
Real numbers encompass all the numbers on the number line, including both rational numbers (like fractions and integers) and irrational numbers (like \( \sqrt{2} \) and \( \pi \)). They provide a complete and continuous scale of values that is used for various measurements and calculations.
In mathematics, when we discuss the domain of functions, we often refer to all real numbers as potential candidates unless specific conditions restrict them.
For instance, in the equation \( f(x) = \frac{1-e^{x^{2}}}{1-e^{1-x^{2}}} \), the domain includes all real numbers except those that result in division by zero. Recognizing where these restrictions occur is crucial to understanding and correctly solving problems involving real numbers.
  • The density of real numbers implies there are infinitely many between any two values.
  • They are foundational in defining continuous functions.
Division by zero
Division by zero is a key mathematical concept that is undefined because it leads to logical contradictions and infinite results. When any number is divided by zero, the result cannot be established within the realm of real numbers, as multiplying zero by any finite number will not re-create the numerator.
Understanding this concept is critical when analyzing functions because it helps determine the limits and proper domains. In function \( f(x) = \frac{1-e^{x^{2}}}{1-e^{1-x^{2}}} \), the expression \( 1 - e^{1-x^2} \) in the denominator can potentially reach zero, leading to undefined values of the function.
Thus, solving \( 1 - e^{1-x^{2}} = 0 \) identifies problematic x-values, specifically \( x = 1 \) and \( x = -1 \).
  • During calculation, always identify which values lead the denominator to zero.
  • Eliminate these values to find the function's domain.

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Most popular questions from this chapter

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Find the exact value of each expression. (a) \(\log _{2} 6-\log _{2} 15+\log _{2} 20\) (b) \(\log _{3} 100-\log _{3} 18-\log _{3} 50\)

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\(\begin{array}{ll}{\text { Use the table to evaluate each expression. }} \\\ {\text { (a) } f(g(1))} & {\text { (b) } g(f(1))} & {\text { (c) } f(f(1))} \\\ {\text { (d) } g(g(1))} & {\text { (e) }(g \circ f)(3)} & {\text { (f) }(f \circ g)(6)}\end{array}\) \(\begin{array}{|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {3} & {1} & {4} & {2} & {2} & {5} \\ \hline g(x) & {6} & {3} & {2} & {1} & {2} & {3} \\ \hline\end{array}\)

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