Domain

Section 1.5 Domain

The set of all possible number (input values), typically \(x\) for which the function is defined and produces a real number output is called a domain. Determining the domain is a fundamental step in analyzing functions, as it identifies the values where the function is valid.

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Steps to Determine the Domain: Identify the function type (polynomial, rational, radical, logarithmic, etc.). Find restrictions: Denominators cannot be zero. Expressions under even roots must be non-negative. Arguments of logarithms must be positive. Express the domain in interval notation or set notation.

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Polynomials: The domain is all real numbers \((\mathbb{R})\) because polynomials are defined for every \(x\text{.}\) Example: For

\begin{equation*} f(x)=x^2+3x+5 \end{equation*}

the domain is \((-\infty, \infty)\text{.}\)

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Rational Functions: Exclude values where the denominator is zero. Example: For

\begin{equation*} f(x) = \frac{1}{x-2} \end{equation*}

solve \(x-2\neq 0\text{,}\) so \(x \neq 2\text{.}\) The domain is \((-\infty, 2) \cup (2, \infty)\text{.}\)

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Square Roots (or Even Roots): The expression inside the root must be non-negative \((\geq 0)\text{.}\) Example: For

\begin{equation*} f(x) = \sqrt{x+3} \end{equation*}

solve \(x+3 \geq 0\text{,}\) so \(x \geq -3\text{.}\) The domain is \([-3, \infty)\text{.}\)

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Logarithms: The argument of a logarithm must be positive \((\gt 0)\text{.}\) Example: For

\begin{equation*} f(x) = \ln(x-1) \end{equation*}

solve \(x-1 \gt 0\text{,}\) so \(x \gt 1\text{.}\) The domain is \((1, \infty)\text{.}\)

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Trigonometric Functions: Most trigonometric functions (e.g., \(\sin(x), \cos(x)\)) have a domain of all real numbers, but others like \(\tan(x)\) or \(\csc(x)\) exclude points where the function is undefined. Example: For

\begin{equation*} f(x) = \tan(x) \end{equation*}

the function is undefined at \(x = \frac{\pi}{2} + k\pi\text{,}\) where \(k \) is an integer. The domain is \(\mathbb{R} \setminus \{ \frac{\pi}{2} + k\pi \mid k \in \mathbb{Z} \}\text{.}\)

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Piecewise Functions: Consider the domain of each piece and combine them, ensuring no overlap unless specified. Example: For

\begin{equation*} f(x)= \begin{cases} x^2 & \text{if } x \geq 0 \\ \frac{1}{x} & \text{if } x \lt 0 \end{cases} \end{equation*}

the domain of \(x^2\) is \(x \geq 0\text{,}\) and the domain of \(\frac{1}{x}\) is \(x \neq 0\text{.}\) Since \(x \lt 0\) for the second piece, the domain is \((-\infty, 0) \cup [0, \infty) = \mathbb{R}\text{.}\)

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Composite Functions: The domain of \(f(g(x))\) requires \(x\) to be in the domain of \(g\text{,}\) and \(g(x)\) to be in the domain of \(f\text{.}\) Example: For

\begin{equation*} f(x) = \sqrt{x} \end{equation*}

and

\begin{equation*} g(x) = x-1\text{,} \end{equation*}

find the domain of \(f(g(x)) = \sqrt{x-1}\text{.}\) Solve \(x-1 \geq 0\text{,}\) so \(x \geq 1\text{.}\) The domain is \([1, \infty)\text{.}\)

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Subsection 1.5.1 Determining Domain:

Example 1.5.1.

\begin{equation*} If \, \quad f(x)=\frac{6}{x} \end{equation*}

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\begin{equation*} If \,\quad f(x) = \sqrt{x} \end{equation*}

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Solution.

we can set \(x=2\) in \(f(x)=\frac{6}{2}=3 \text{.}\) Hence, \(x=2\) is a domain which defines \(f(x) =3 \) in a real number. Again, set \(x=0\) in \(f(x)=\frac{6}{0} = \) undefined or \(\infty\text{.}\) Hence, \(x=0\) is a not a domain of the given \(f(x)=\infty\) in a real number.
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we can set \(x=4\) in \(f(x)=\sqrt{4} = 2 \text{.}\) Hence, \(x=4\) is a domain which defines \(f(x)=2\) in a real number. Again, set \(x=-1\) in \(f(x)=\sqrt{-1} = \) imaginary. Hence, \(x=-1\) is a not a domain of the given \(f(x)\) in a real number.
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Domain is any input values that defines a function in real number.

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Example 1.5.2.

Type I: Only Polynomials.

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\begin{equation*} f(x) = 4x+1 \end{equation*}

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Solution.

Let \(x=2\text{,}\) then \(f(x) = 4\times 2+1 = 9\) (defined), hence \(x=2\) is a domain of a \(f(x).\)

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let \(x=-1\text{,}\) then \(f(x) = 4\times -1+1 = -3\) (defined), hence \(x=-1\) is a domain of a \(f(x).\)

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i.e., we can put any value of a domain from real numbers to define the given function \(f(x).\)

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Domain: \((-\infty, +\infty)\) or all real numbers.

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Example 1.5.3.

Type II: Only Fractions.

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\begin{equation*} f(x) = \frac{3x+6}{x^2+3x+2} \end{equation*}

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Solution.

Here, denominator \(\neq 0\text{,}\) otherwise the function gets undefined.

\begin{equation*} \therefore \, x^2+3x+2 \neq 0 \end{equation*}

\begin{equation*} or, \, (x+2)(x+1)\neq 0 \end{equation*}

\begin{equation*} or, \, x \neq -2, -1 \end{equation*}

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Domain: \((-\infty, -2)\cup (-2, -1)\cup (-1, \infty)\) or all real numbers except \(x=-1,-2.\) Because these numbers make denominator \(=0.\)

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Example 1.5.4.

Type III: Square Root on Bottom.

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\begin{equation*} f(x) = \frac{x^2+6x+9}{\sqrt{x+3}} \end{equation*}

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Solution.

Here, denominator \(\gt 0\text{,}\) otherwise the function gets undefined.

\begin{equation*} or, \text{ radicand } \, x+3 \gt 0 \end{equation*}

\begin{equation*} or, \, x \gt -3 \end{equation*}

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\(x+3 =0 \) makes \(f\) undefined, and \(x+3 \lt 0\) makes \(f\) imaginary.

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Domain: \((-3,\infty)\) or \(\forall_x \gt -3.\)

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Example 1.5.5.

Type IV: Square Root on Top.

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\begin{equation*} f(x) = \frac{\sqrt{x-1}}{x^2-4} \end{equation*}

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Solution.

Here, denominator \(\neq 0\) and radicand \(\geq 0\)

\begin{equation*} or, \, x^2-4 \neq 0 \end{equation*}

\begin{equation*} or, \, (x+2)(x-2) \neq 0 \, \Rightarrow \, x \neq -2,+2 \end{equation*}

\begin{equation*} and \, \quad x-1 \geq 0 \, \Rightarrow \, x \geq 1 \end{equation*}

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Domain: \([1,2) \cup (2, \infty) \) or all real numbers of \(x \geq 1\) excluding \(x= 2.\)

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Example 1.5.6.

Type V: Only Square Root.

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\begin{equation*} f(x) = \sqrt{x^2-x-12} \end{equation*}

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Solution.

Here, radicand \(\geq 0\)

\begin{equation*} or, \, x^2-x-12 \geq 0 \end{equation*}

\begin{equation*} or, \, (x-4)(x+3) \geq 0 \end{equation*}

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set \(x=5\text{,}\) \(x=2,\) and \(x=-4,\) in regions \(A, B,\) and \(C\) respectively, as shown in graph.

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Check:

A region: at any value \(x \gt 4\) e.g., \(x=5\)

\begin{equation*} (x-4)(x+3) = (5-4)(5+3) = 8 =+ve \end{equation*}

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B region: at any value \(x \gt -3\) e.g., \(x=2\)

\begin{equation*} (x-4)(x+3) = (2-4)(2+3) =-10 =-ve \end{equation*}

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C region: at any value \(x \lt -3\) e.g., \(x=-4\)

\begin{equation*} (x-4)(x+3) = (-4-4)(-4+3) =8 =+ve \end{equation*}

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Since, \(-ve \) value of radicand is not allowed as we cannot have \(\sqrt{-ve}\) number. our domain looks like: Domain: \((-\infty,-3] \cup [4, \infty) \text{.}\)

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Example 1.5.7.

Find the domain of the following functions:

\begin{equation*} f(x) = 3x-2 \end{equation*}

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\begin{equation*} f(x) = x^2+5x-11 \end{equation*}

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\begin{equation*} f(x) = 2x^3-x^2+3x-5 \end{equation*}

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\begin{equation*} f(x) = 6x^7+5x^5-4x-10 \end{equation*}

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\begin{equation*} \sqrt[3]{8} = \sqrt[3]{2^3} = 2 \, \text{ Real}. \end{equation*}

or,

\begin{equation*} \sqrt[3]{-8} = \sqrt[3]{(-2)^3} = -2 \, \text{ Real} \end{equation*}

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If \(n\) is even, then \(expression \, \geq 0\text{.}\) The domain of radical functions with an even index and expression \(\geq 0\text{,}\) is all real numbers.

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For example:

\begin{equation*} \sqrt[4]{16} = \sqrt[4]{2^4} = 2 \, \text{ Real} \end{equation*}

or,

\begin{equation*} \sqrt[3]{-16} = \sqrt[4]{(-2)^4} = 2i \, \text{ not Real} \end{equation*}

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Example 1.5.12.

\begin{equation*} f(x)=\sqrt{x-9} \end{equation*}

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Solution.

Here, \(n\) is even, hence, set,

\begin{equation*} x-9 \geq 0 \rightarrow \, x \geq 9 \end{equation*}

Hence, Domain, \(\mathbb R; [9, \infty).\)

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Example 1.5.13.

\begin{equation*} f(x)=\sqrt[4]{6-2x} \end{equation*}

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Solution.

Here, \(n\) is even, hence, set,

\begin{equation*} 6-2x \geq 0 \rightarrow \, 6 \geq 2x \end{equation*}

or,

\begin{equation*} x \leq 3 \end{equation*}

Hence, Domain, \(\mathbb R; (-\infty, 3].\)

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Example 1.5.14.

\begin{equation*} f(x)=\sqrt{x^2+4x-21} \end{equation*}

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Solution.

Here, \(n\) is even, hence, set,

\begin{align*} x^2+4x-21 \amp \geq 0 \\ x^2+7x-3x-21 \amp \geq 0 \\ (x-3)(x+7) \amp \geq 0 \end{align*}

Hence,

\begin{equation*} x-3 \geq 0 \rightarrow \, x \geq 3 \end{equation*}

and

\begin{equation*} x+7 \geq 0 \rightarrow \, x \geq -7 \end{equation*}

but, for \(x \geq -7\) gives function expression negative and for \(x \leq -7\) gives positive expression. Hence, we concluded \(x \leq -7\text{.}\) Hence, Domain, \(\mathbb R; (-\infty, -7] \cup [3, \infty).\) If \(x=0\text{,}\) \(f(x) \) becomes square root of negative expression and same will be seen if \(x \geq -7.\)

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