Finding Domain of a Function
Domain of a Function
The process of determining the domain of a function involves identifying the largest subset of \( \mathbb{R} \) where the function is well-defined. This requires remembering the domains of elementary functions and understanding how the algebra of functions influences domains. By systematically applying these principles, we can calculate the domain of any given function.
In this section, we focus on domain calculation. Specifically, we consider the natural domain, which is the largest subset of \( \mathbb{R} \) such that the expression defining the function is mathematically valid. This involves excluding values that lead to undefined operations such as division by zero, square roots of negative numbers, or logarithms of non-positive values.
You should revisit the domain of fundamental functions, such as polynomials, rational functions, square root functions, logarithmic functions, and trigonometric functions, as these serve as the building blocks for more complex expressions. Additionally, it is important to understand how operations like addition, subtraction, multiplication, division, and composition affect the domain, as these operations often introduce new constraints or restrictions that must be accounted for when calculating the domain of combined or derived functions.
We begin with some common patterns in mathematical expression formed using elementary fucntions and algebraic operations.
Square Root
For a function \( f(x) = \sqrt{g(x)} \), where \( g(x) \) is any expression in \( x \), the square root imposes the restriction that \( g(x) \geq 0 \). This restriction arises from the fact that the square root function is undefined for negative arguments in the set of real numbers. To determine the domain of \( f(x) \), we solve the inequality \( g(x) \geq 0 \), which provides the set of all \( x \) for which \( f(x) \) is well-defined.
Consider \( f(x) = \sqrt{x+2} + \sqrt{1-x} \). This function comprises two terms, each imposing its own condition on the domain:
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The first term, \( \sqrt{x+2} \):
The square root function requires \( x+2 \geq 0 \). Solving this inequality:
\[ x \geq -2 \] -
The second term, \( \sqrt{1-x} \):
Similarly, the square root function requires \( 1-x \geq 0 \). Solving this inequality:
\[ x \leq 1 \]
The domain of \( f(x) \) is the set of \( x \) that satisfies both conditions simultaneously, as both square roots must be defined. This means the domain is given by the intersection of the intervals:
Combining these, we obtain:
Thus, the domain of \( f(x) \) is determined to be:
Example
Find the domain of \( h(x) = \sqrt{(x+1)(x-2)} - 3\sqrt{|x-1|-2} \).
Solution:
To determine the domain of \( h(x) \), we analyze the constraints imposed by each term.
The first term, \( \sqrt{(x+1)(x-2)} \), requires \( (x+1)(x-2) \geq 0 \) since the square root function is only defined for non-negative arguments. Solving the inequality \( (x+1)(x-2) \geq 0 \) using sign analysis, the product is non-negative in the intervals \( x \in [-1, 2] \cup [2, \infty) \). Thus, the domain for this term is \( x \in [-1, \infty) \).
The second term, \( \sqrt{|x-1|-2} \), requires \( |x-1|-2 \geq 0 \). Simplifying this inequality, we get \( |x-1| \geq 2 \), which splits into two cases:
yielding \( x \geq 3 \) or \( x \leq -1 \). Thus, the domain for this term is \( x \in (-\infty, -1] \cup [3, \infty) \).
Finally, for \( h(x) \) to be defined, both constraints must hold simultaneously. Taking the intersection of the domains \( x \in [-1, \infty) \) and \( x \in (-\infty, -1] \cup [3, \infty) \), we find that the domain of \( h(x) \) is:
Radicals in General
For even radicals such as \( \sqrt[2n]{g(x)} \), where \( n \in \mathbb{N} \), the expression is defined only when \( g(x) \geq 0 \). This is because even roots of negative numbers are undefined in the real number system. Solving \( g(x) \geq 0 \) provides the domain of the expression.
On the other hand, for odd radicals such as \( \sqrt[3]{g(x)} \), the expression is always defined for any real value of \( g(x) \). Odd roots do not impose any restrictions on \( g(x) \), as the cube root of a negative number (or any odd root) is also a well-defined real number. Therefore, \( \sqrt[3]{g(x)} \) is valid for all \( g(x) \in \mathbb{R} \).
Reciprocal Function
For a function of the form \( f(x) = \frac{1}{g(x)} \), the expression is undefined when \( g(x) = 0 \), as division by zero is not defined in the real number system. The domain of \( f(x) \) is therefore all \( x \in \mathbb{R} \) except those for which \( g(x) = 0 \).
Consider \( f(x) = \frac{1}{x-1} \). Here, \( g(x) = x-1 \). The restriction \( g(x) \neq 0 \) translates to:
Thus, the domain of \( f(x) \) is:
In interval notation, the domain is:
Logarithmic Function with Variable Base and Argument
For a function of the form \( \log_{f(x)}(g(x)) \), where both the base \( f(x) \) and the argument \( g(x) \) depend on \( x \), the domain is determined by the following conditions:
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The base \( f(x) \) must satisfy:
\[ f(x) > 0 \quad \text{and} \quad f(x) \neq 1. \] -
The argument \( g(x) \) must satisfy:
\[ g(x) > 0. \]
These conditions ensure the logarithmic function is well-defined.
Example: \( f(x) = \log_{1-x}(x+3) \)
Here, the base is \( 1-x \) and the argument is \( x+3 \). The domain is determined by solving the following inequalities:
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Condition on the base \( 1-x \):
\[ 1-x > 0 \implies x < 1, \]\[ 1-x \neq 1 \implies x \neq 0. \] -
Condition on the argument \( x+3 \):
\[ x+3 > 0 \implies x > -3. \]
Combining \( x < 1 \), \( x \neq 0 \), and \( x > -3 \), we find the domain is:
In interval notation, the domain of \( f(x) \) is:
Domain of f(x) = tan(g(x))
The tangent function is undefined wherever its argument equals \( (2n+1)\frac{\pi}{2} \), where \( n \in \mathbb{Z} \). For \( f(x) = \tan(g(x)) \), this translates to the condition:
Domain Determination
To find the domain of \( f(x) \), exclude all \( x \) such that \( g(x) = (2n+1)\frac{\pi}{2} \). Hence, the domain of \( f(x) \) is:
This ensures \( f(x) \) remains well-defined, avoiding singularities of the tangent function.
Domain of f(x) = \cot(g(x))
The cotangent function is undefined wherever its argument equals \( n\pi \), where \( n \in \mathbb{Z} \). For \( f(x) = \cot(g(x)) \), this translates to the condition:
Domain of \( \sin(f(x)) \) and \( \cos(f(x)) \)
The functions \( \sin(f(x)) \) and \( \cos(f(x)) \) are always defined for all real values of \( f(x) \). Unlike other trigonometric functions like \( \tan(f(x)) \) or \( \cot(f(x)) \), there are no restrictions on \( f(x) \), as both \( \sin(x) \) and \( \cos(x) \) are defined for all \( x \in \mathbb{R} \).
Thus, for both cases:
Domain Determination
To find the domain of \( f(x) \), exclude all \( x \) such that \( g(x) = n\pi \). Hence, the domain of \( f(x) \) is:
This ensures \( f(x) \) remains well-defined, avoiding the singularities of the cotangent function.
Domain of Inverse Trigonometric Functions
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Domain of \( \sin^{-1}(f(x)) \):
The arcsine function is defined only when \( -1 \leq f(x) \leq 1 \). Thus, the domain is:
\[ \{x : -1 \leq f(x) \leq 1\}. \] -
Domain of \( \cos^{-1}(f(x)) \):
The arccosine function has the same restriction as the arcsine function. Hence, the domain is:
\[ \{x : -1 \leq f(x) \leq 1\}. \] -
Domain of \( \sec^{-1}(f(x)) \):
The arcsecant function is defined only when \( |f(x)| \geq 1 \). Therefore, the domain is:
\[ \{x : |f(x)| \geq 1\}. \] -
Domain of \( \csc^{-1}(f(x)) \):
The arccosecant function has the same restriction as the arcsecant function. Thus, the domain is:
\[ \{x : |f(x)| \geq 1\}. \]
Domain of \( \tan^{-1}(f(x)) \) and \( \cot^{-1}(f(x)) \)
The functions \( \tan^{-1}(f(x)) \) and \( \cot^{-1}(f(x)) \) are always defined for all real values of \( f(x) \). Unlike other inverse trigonometric functions, there are no restrictions on \( f(x) \), as the domains of both \( \tan^{-1}(x) \) and \( \cot^{-1}(x) \) cover all real numbers.
Thus, for both cases: