For a function $f: A \rightarrow B$
Set A is called the domain of the function f
Set B is the called the codomain of the function
For real function, A and B are subset of the real numbers.
In some cases,domain of the real function may not be explicity defined. We are just given the function
y=f(x)
In such cases domain would mean the set of real values of x for which f(x) assumes real values.Domain of the function f is denoted by D(f)
Therefore,
D(f) = {$x \in R$, f(x) is a real number}
Example
1. $y=x^2$ Solution
Here it is clear that y assumes real values for all $x \in R$
So, D(y) =R
2. Let A= {2,4,5,6}. A function f is defined from X to N
$f={(x,y),y=x^3 + 2, x \in A ,y \in N }$
Find the domain and codomain of the function? Solution
Domain will be Set A
D(f) ={2,4,5,6}
Codomain will be N
How to find the domain of a function
Find the values of x for which given real function takes the below forms
a. $ \frac {1}{0}$
b. $ \sqrt {a \; negative \; integer}$
c. $ \frac {1}{\sqrt {a \; negative \; integer}}$
d. $ \frac {0}{0}$
e. $ log (a negative integer)$
f. $ \frac {1} {log (a \; negative \; integer) \; or (\; equal \; to \; 1)}$
The set of real numbers excluding those obtained in step 1 is the domain of the function
How to write the domain in interval form
\((a,b)\)
It is the open interval set between point and b such that All the points
between a and b belong to the open interval (a, b) but a, b themselves do not belong to this interval
\(\{ y:a < y < b\} \)
\([a,b]\)
It is the closed interval set between point and b such that All the points
between a and b belong to the open interval (a, b) including a, b
\(\{ x:a \le x \le b\} \)
\([a,b)\)
It is the open interval set between point and b such that All the points
between a and b belong to the open interval (a, b) including a, but not b
\(\{ x:a \le x < b\} \)
\((a,b)\)
It is the open interval set between point and b such that All the points
between a and b belong to the open interval (a, b) including b, but not a
\(\{ x:a < x \le b\} \)
Solved Examples
Find the domain of the below functions
1. $y =f(x) = \frac {1}{x-1}$ Solution
Here we see that function is defined for all values of $x \in R$ except for x=1 (it becomes of the form 1/0 for x=1) .
So domain will be given
D(f) = R -1
2. $f(x) = \frac {1}{ \sqrt {x -3}}$ Solution
The function is defined only
x -3 > 0 or x > 3
Domain will be given as
D(f) =$(3,\infty )$
3. $f(x) = \frac {1}{x^2 + 3}$ Solution
Since $x^2 + 3$ > 0 for $x \in R$, This function is defined for all values of $x \in R$
D(f) =$(- \infty,\infty )$
4.$ f(x) = \sqrt {(x-3)(x-4)}$ Solution
Here for f(x) to be real,
$ (x-3)(x-4) \geq 0 $
or $ x \in (-\infty ,3] \cup [4,\infty]$
So domain of the function will be
D(f) = $(-\infty ,3] \cup [4,\infty]$