A set of elements for which a function has been defined is called the domain of a function.

Every function is defined for some elements by giving their associated values. The collection of such elements as a group is known as domain of the function.

Each function has a domain and it helps us to understand the list of values for which a function is defined mathematically.

Consider a function and assume it is defined for the elements $a$, $b$, $c$, $d$ and $e$. Take $x$ as a literal and it represents all these elements. Assume, the function is expressed as $f(x)$ in mathematics.

Take $x = a$, then the value of the function is $f(a)$

Take $x = b$, then the value of the function is $f(b)$

Take $x = c$, then the value of the function is $f(c)$

Take $x = d$, then the value of the function is $f(d)$

Take $x = e$, then the value of the function is $f(e)$

The values of function are $f(a)$, $f(b)$, $f(c)$, $f(d)$ and $f(e)$ for $x$ is equal to $a$, $b$, $c$, $d$ and $e$ respectively. The collection of the elements $a$, $b$, $c$, $d$ and $e$ is called the domain of the function.

$x = \{a, b, c, d, e\}$

$f(x) = \dfrac{1}{x+2}$

Substitute all the real numbers to determine the domain of this function.

For example, Take $x = 0$ and evaluate the function.

$x = 0$ $\implies$ $f(0) = \dfrac{1}{0+2} = \dfrac{1}{2}$

For $x$ is equal to zero, the value of the function is $\dfrac{1}{2}$. Now, substitute positive real numbers and obtain their associated values.

$x = 1$ $\implies$ $f(1) = \dfrac{1}{1+2} = \dfrac{1}{3}$

$x = 2$ $\implies$ $f(2) = \dfrac{1}{2+2} = \dfrac{1}{4}$

$x = 3$ $\implies$ $f(3) = \dfrac{1}{3+2} = \dfrac{1}{5}$

$x = 4$ $\implies$ $f(4) = \dfrac{1}{4+2} = \dfrac{1}{6}$

$x = 5$ $\implies$ $f(5) = \dfrac{1}{5+2} = \dfrac{1}{7}$

$\vdots$

The function is defined for all the positive real numbers.

Now, substitute all the negative real numbers one by one and test the functionality of this function.

$x = -1$ $\implies$ $f(-1) = \dfrac{1}{-1+2} = \dfrac{1}{1} = 1$

$x = -2$ $\implies$ $f(-2) = \dfrac{1}{-2+2} = \dfrac{1}{0} = \infty$

For the value $x$ is equal to $-2$, the example function becomes undefined. Hence, the value of $x$ should not be equal to $-2$. Continue testing the function by substituting the remaining negative real numbers.

$x = -3$ $\implies$ $f(-3) = \dfrac{1}{-3+2} = \dfrac{1}{-1} = -1 $

$x = -4$ $\implies$ $f(-4) = \dfrac{1}{-4+2} = \dfrac{1}{-2} = -\dfrac{1}{2}$

$x = -5$ $\implies$ $f(-5) = \dfrac{1}{-5+2} = \dfrac{1}{-3} = -\dfrac{1}{3}$

$\vdots$

Now, collect all the elements as a set but ignore $x$ is equal to $-2$. Therefore, the collection of elements is called the domain of this function.

$x = \{\cdots -4, -3, -1, 0, 1, 2, 3, \cdots\}$

According to the set theory, the real numbers group is simplify denoted by $R$ but it contains $-2$. Hence, it should be subtracted from it. Therefore, the domain of the function is simplify written as $R-\{0\}$.

List of most recently solved mathematics problems.

Jul 04, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

Jun 23, 2018

Limit (Calculus)

Evaluate $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{e^{x^2}-\cos{x}}{x^2}$

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.