 # Range of a Function

## Range of a Function

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}

### Quiz Time

Question 1 Find the range of the function $f(x) = \frac {x-5}{x-3}$
A. R -{5/3}
B. R -{1}
C. R -{-1}
D. None of the above
Question 2Find the range of the function $f(x) = \sqrt {x-1}$ ?
A. $(1,\infty )$
B. $(0,\infty )$
C. $[0,\infty )$
D. $[1,\infty )$
Question 3 if function f : X -> R, $f (x) = x^3$, where X = {-1, 0, 3, 9, 7}, The Range of the function will be
A. {-1,0,3,9,7}
B. {-1,1,9,49,81}
C. {-1,0,27,343}
D. {-1,0,27,343,729}
Question 4 Let f(x) =x2 ,find the value of $\frac {f(2.1) -f(2)}{2.1 -2}$
A. 4
B. .41
C. 4.1
D. .14
Question 5 Find the range of the function defined as $f(x)=\sqrt {4-x^2}$
A. [-2,2]
B. [0,2]
C.(0,2)
D.[-2,0)
Question 6Find the Range and domain of the function $f(x) =\frac {x -11}{22-2x}$
A. Domain = R, Range = {-1/2, 1/2}
B. Domain = R - {1}, Range = R
C. Domain = R - {11}, Range = {-1/2}
D. Domain = R - {� 11}, Range = {-1/2, 1/2}

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