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Learning objectives:

This article covers the steps involved in finding the domain of any function f(x).

Following are the concepts covered and features of this article

  • What is the function?
  • Polynomial Functions & Rational functions.
  • The domain of a function.
  • Finding the restrictions on “x” values for the domain of a  function.
  • Steps to find the domain of a function.

What is the function?

A function is generally represented by y= f(x).

Here the  “x” is called an independent variable and the “y” is the dependent variable.

Every point on the function is represented by (x,y) coordinates.

A relation between “x” and “y” variables is a function if and only if every “x” has a different corresponding “y” value.

NO Two distinct “y” values must map to the same  “x” value.

A vertical line test is generally used to identify whether a relation is a function.

An easy way to identify a function is “LOOK FOR A EVEN POWER OF “Y”.If we find the even power of “y”, then it is not a  function. For example  Y^2= 4x, Y^4 = 4x   have the even power of  Y.

Functions with even power of “Y” fail the vertical line test.


Here we can observe that the x=4  is mapped to 4 and -4  as  well.

Polynomial Functions & Rational functions:

Polynomial functions  of the form  f\left( x\right) =x^{n}+x^{n-1}+x^{n-2}+ -..\ldots +k.

The exponents of “x” always positive. The Negative exponents for  “x” are not allowed.

There will be no fractional terms in the polynomial functions.

On the other hand, rational functions are of  P/Q  from and there will be  “x” in the denominator.

f\left( x\right) =\dfrac {x^{2}+1}{x+2}.


The domain of a function.

We know that divide by zero  ( say 1/0)  is undefined and the square root of a negative number is a complex number which is generally represented by “i”.

For any function, we need to ensure that we do not get the above two cases.

The domain of a function is all about choosing the permissible  “x” values such that, the function does not land up in the above mentioned special cases of divide by zero or the square root of a negative number.

The Domain of a function is the set of all “PERMISSIBLE” values of the independent variable “x”.

Finding the restrictions on “x” values for the domain of a  function.

As discussed above, to find the permissible “x” values of a function, we need to find the  Restrictions on the function.

Restrictions are the values of “x”  that the function is NOT allowed to take.

Examples for different types of functions and restrictions of a function:

i)f\left( x\right) =x

  • In this function, we observe that there is no fraction and there is no square root. So  there will  be no restrictions for choosing the “x” for this function. As a result ,we can say that the  domain of the  function is  All real Numbers and X  \in R . \left( -\infty ,\infty \right) is the interval notation.

ii) f\left( x\right) =\dfrac {1}{x+2}

  • This function has a denominator and there is a chance of getting a “Divide by ‘0’  case”.So to avoid it, we ensure that the  denomiantor part   x+2 \neq 0 . Hence    X\neq -2
  • We are allowed to take any value for  “X” but not the x =-2. So x =-2  is the restriction on choosing “x”  for  this function.
  • Hence The domain of function can be represented as \left( -\infty ,-2\right) \cup \left( -2,\infty \right) OR  R- {-2}.

iii)\sqrt {x-3}

  • The function above does not have a  denominator.
  • Therefore, there is no possibility to get a ” divide by 0″ case.
  • However, the function has got a square root and so, there is a  possibility of  “Square root of a negative number” case.
  • To avoid square root negative  number we can write the in-equality                                                                                                                         \begin{aligned}x-3\geq 0\\ \Rightarrow x\geq 3\end{aligned}
  • As a result, the  domain of  the  function would be   x\geq 3 OR  \left[ 3,\infty \right).

iv) \dfrac {1}{\sqrt {x-5}}

  • This function has both denominator and square root.
  • So as the first step, we check for the restriction on the “x” to avoid divide by 0.
  • Therefore                                                                                                                                                                                                                                  \begin{aligned}\sqrt {x-5}=0\\ \Rightarrow x-5=0\\ \Rightarrow x=5\end{aligned} .The X  must not be equal to 5.
  • We have a square root for X-5. Therefore to avoid “square root of a negative  number”                                              \begin{aligned}x-5\geq 0\\ x\geq 5\end{aligned}.
  • Since we are not supposed to take  x=5, the final solution would be x>5 only.
  • As a result, The domain of function is  X>5 OR \left( 5,\infty \right) .
  • The only change from the  previous function \sqrt {x-3} domain is that “=” sign will not be taken along with  > symbol

v) \dfrac {\sqrt {x-5}}{x+3}

  • This  rational function  has   x+3 denominator  and  a square root  function \sqrt {x-5}  in the  numerator.
  • So we need to find the restrictions for each case and take the common solution which works for both.
  • The denominator is x+3. So we must take x+3 =0. Hence X=-3 and we are NOT supposed to take x =-3
  • For the  numerator, we  have \sqrt {x-5}.To avoid square root negative number  we take                                  \begin{aligned}x-5\geq 0\\ \Rightarrow x\geq 5\end{aligned}.
  • So  the  solution which is common for both x\neq -3and  x\geq 5 is x\geq 5.
  • Therefore the  final Domain of the  function is x\geq 5 OR \left[ 5,\infty \right).

vi) \dfrac {1}{\sqrt {x^{2}+6}}

  • This rational function has an  x^2+6 in the denominator.
  • In order to find the  restriction on this function, we equate it to 0.                                                                                              \begin{aligned}\sqrt {x^{2}+6}\neq 0\\ \Rightarrow x^{2}+6\neq 0\\ x^{2}\neq -6\\ x\neq \sqrt {-6}\end{aligned}
  • Here we observe that the  x^2 term can never become a negative  value.The  least value x^2 can take is 0 only.
  • Therefore The domain of function is All real number R. In the interval notation, \left( -\infty ,\infty \right) is the solution.

vii) \dfrac {\sqrt {x+3}}{\sqrt {\left( x+6\right) \left( x-7\right) }}

  • Given a rational function having square roots in the numerator and denominator as well.
  • So the strategy is that we find the TWO restrictions for the square root in the numerator and the denominator and look for the individual restrictions.
  • Once we find the respective individual restrictions, we find a common solution that works for both restrictions.
  • For the numerator, the  function inside  the  square root  must be  >= 0                                                                                                                   \begin{aligned}x+3\geq 0\\ \Rightarrow x\geq -3\end{aligned}.
  • For the denominator part, the (x+6)(x-7) must be greater than 0. Therefore for the denominator to be positive, either both  x+6>0 and  x-7>0 OR  x+6<0 and x-7<0.
  • Hence X>-6 and x>7  OR x<-6 and x<7. The solutions  x<-6 and x<7 are ruled out since x must be  greater than -3 from earlier restriction.
  • Therefore the consolidated domain that is common for x>-3,x>-6, and x>7  is x>7 itself.
  • So ,The domain of function would be X >7 .The interval notation is  \left( 7,\infty \right) .



The domain of a function is all Real numbers R  if there is no denominator and No square root in the function.

   If the function has a denominator.

  • If there is a denominator, then equate the denominator = “0” to get the restrictions for the Domain of the function.                                                                                                                                                                                                                                                                                      \begin{aligned}\dfrac {1}{x-a}\\ \Rightarrow x-a\neq 0\\ x\neq a\\ Domain:\left( -\infty ,a\right) \cup \left( a,\infty \right) \end{aligned}.


  • If there is a denominator with a”square root” write the “in-equality” making denominator >0 for the Domain of the function.                                                                                                                                                                                                                                                      \begin{aligned}\dfrac {1}{\sqrt {x-a}}\\ \Rightarrow \sqrt {x-a} >0\\ \Rightarrow x-a >0\\ \Rightarrow x >a\\ Domain:\left( a,\infty \right) \end{aligned}.

   If the function has got a numerator and no denominator.

  • If there is numerator alone, then we check whether there is a square root for the numerator.
  • If there is a square root in the numerator , we make “In -Equality” >=0 .                                                                        \begin{aligned}\sqrt {x+a}\\ \Rightarrow \sqrt {x+a}\geq 0\\ \Rightarrow x+a\geq 0\\ \Rightarrow x\geq -a\\ Domain:\left[ -a,\infty \right) \end{aligned}
  • Notice that here we have the in-equality >=0, Unlike earlier denominator case which is having in-equality “>0″since the numerator is allowed to take a zero.


If the function has both Numerator & Denominator.

  • If the function has both numerator and denominator, we find the individual restrictions for the numerator and the denominator as mentioned in the above steps.
  • We take the common solution for both numerator and the denominator.

\begin{aligned}\dfrac {\sqrt {x+5}}{x+4}\\ \Rightarrow x+4\neq 0,x+5\geq 0\\ \Rightarrow x\neq -4,x\geq -5\\ Domain=\left[ -5,-4\right) \cup \left(- 4,\infty \right) \end{aligned}.

  • Here we see that we are allowed to take all values of x >=-5  EXCLUDING   the x =-4.