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
Set of Images of all elements in Set A is called the range i.e it is the set of values of f(x) which we get for each and every x in the domain
It is obvious that range could be subset of co-domain as their may be elements in co-domain which are not the images of any element in domain
It is denoted by R( f) or Range (f)
For real function, A and B are subset of the real numbers.
Example
Find the domain and Range of
1. Find the domain and Range of the function given by $y=x^2$ Solution
Here it is clear that y assumes real values for all $x \in R$
So, D(y) =R
Now we can see that y value is always positive or zero i.e $y \geq 0$
So Range of function will be
Range (f) = $[0,\infty)$
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 Range of the function? Solution
D(f) ={2,4,5,6}
Let calculate the value of the function for each values of x in domain
for $x=2 , y=x^3 + 2 = 2^3 +2 = 10$
for $x=4 , y=x^3 + 2 = 4^3 +2 = 66$
for $x=5 , y=x^3 + 2 = 5^3 +2 = 127$
for $x=6 , y=x^3 + 2 = 2^3 +2 = 218$
So, Range of the function will be given by
R(f) ={10,77,127,218}
How to find the Range of a function
There are many method to find the range of a function
A.Range of the function may be find using below algorithm. This is inverse function technique
put y=f(x)
Solve the equation y=f(x) for x in terms of y ,let x =g(y)
Find the range of values of y for which the value x obtained are real and are in the domain of f
The range of values obtained for y is the Range of the function
B.One way would be checking the function for different value of x and drawing the graph and then draw the conclusion for Range
C.Another method would be to look for minimum and maximum values of function and then find the range
How to write the Range 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 and Range of the below functions
1. $y =f(x) = \frac {2}{x-1}$ 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}
For Range , let y =f(x) i.e
$ y=\frac {2}{x-1}$
$y(x-1)=2$
$yx -y=2$
$yx=2+y$
$x =\frac {2+y}{y}$
Obviously x assumes real value for all y except y=0
Hence Range is R(f) = R -{0}
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) =$(3,\infty )$
For Range , let y =f(x) i.e
$y = \frac {1}{ \sqrt {x -3}}$
$y^2 = \frac {1}{x-3}$
$xy^2 -3y^2 =1$
$x= \frac {1 +3y^2}{y^2}$
Obviously x assumes real value for all y except y=0
Also $y=\frac {1}{ \sqrt {x -3}} > 0$
Therefore,Range of function R(f) =$(0,\infty)$
3. $f(x) = \frac {x}{x^2 + 3}$ Solution
Since $x^2 + 3$ > 0 for $x \in R$, This function is defined for all values of $x \in R$
D(f) =$(- \infty,\infty )$
For Range , let y =f(x) i.e
$y = \frac {x}{x^2 + 3}$
$yx^2 + 3y -x =0$
clearly for y=0,x=0
Now for $y \neq 0$
$x = \frac {1 \pm \sqrt {1 -12y^2}}{2y}$
Obviously, x will resume real values if
$1 -12y^2 \geq 0$
$y^2 - \frac {1}{12} \leq 0$
or $ -\frac {1}{2\sqrt 3} \leq y \leq \frac {1}{2\sqrt 3}$
Hence Range of y is
R(f) = $[-\frac {1}{2\sqrt 3},\frac {1}{2\sqrt 3}]$
4.$ f(x) = \frac {(x-3)}{(3-x)}$ Solution
$f(x) = \frac {(x-3)}{(3-x)}$
or
$f(x) = -\frac {(x-3)}{(x-3)}$
We can observe that this assumes real values for all values of $x \in R$ except 0( it becomes of the undefined form 0/0).So Domain of the function is R -{3}
Now for values of x in domain, this function can be written as
$f(x) =-1$
Clearly Range of the function is {-1}
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