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For a function
f:A ->B
Set A is called the domain of the function f
Set B is called the co-domain of the function
The set of Images of all elements in Set A is called the range i.e it is the set of values of f(x) that we get for each and every x in the domain
Now let’s see how to find the range of a function algebraically i.e without plotting the graph
How to find the range of a function algebraically
General Method is explained below. This is called the inverse function technique
(a) put y=f(x)
(b) Solve the equation y=f(x) for x in terms of y ,let x =g(y)
(c) Find the range of values of y for which the value x obtained are real and are in the domain of f
(d) The range of values obtained for y is the Range of the function
This is basically how to find the range of a function without graphing
Let’s see fee examples with various types of functions
How to find the range of a rational function
a. $f(x) = \frac {x-3}{x-1}$
First, let’s see the domain of the function
We can see that function is defined for all values of x except 1
So Range is $R -{1}$
Now let’s find the range using the inverse function method
$y = \frac {x-3}{x-1}$
$y(x-1) = x-3$
$xy -y = x-3$
$3-y = x -xy$
$x= \frac {3-y}{1 -y}$
It is very clear that x assumes real values for all y except y=1, So Range is
$R – {1}$
b. $f(x) = \frac {x^2 -16}{x-4}$
First let’s see the domain
We can see that function is defined for all values of x except 4
So Range is $R -{4}$
Now $f(x) = \frac {x^2 -16}{x-4}$
$f(x) = \frac {(x-4)(x+4)}{x-4}$
$f(x) = x+4$
or y =x+4
x= y -4
It is very clear that x assumes real values for all y
But there is one catch, we got this equation only when $x \ne 4$, so y=8 would not be in the range of the function.
So,Range is $R – {4}$
How to find the range of a quadratic function/polynomial function
A quadratic Function /Polynomial function is like $f(x) = ax^2 + bx +c $.
a. $f(x) =x^2 -1 $
Clearly, it is defined for all values of x, Domain =R
Now
$y =x^2 -1$
$x^2 = y+1$
$x = \pm \sqrt {y+1}$
For x to be real , $y +1 \geq 0$ or $ y \geq -1$
So $Range = [ -1 , \infty )$
b. $f(x) =-2x^2 -1 $
Clearly, it is defined for all values of x, Domain =R
Now
$y =-2x^2 -1$
$y +1 =-2x^2$
$ x^2 =\frac {-1-y}{2}$
$x = \pm \sqrt {\frac {-1-y}{2}}$
For x to be real , $\frac {-1-y}{2} \geq 0 $
or $ -1 -y \geq 0$
$ -1 \geq y$
$ y \leq -1$
So $Range = (\infty , -1]$
How to find the range of a modulus function
A modulus Function is like $f(x) = |x-1|$
a. $f(x) = 1 – |x-3|$
Clearly, it is defined for all values of x, Domain =R
Now
$y= 1 -|x-3|$
$|x-3| = 1- y$
Clearly for real values of x, $1-y \geq 0$
or
$ 1 \geq y$
So Range is $(-\infty , 1]$
Similarly, we find the range of many functions algebraically i.e without plotting the graph. (how to find range of a function without graphing)
Some Practice Questions
a. $ f(x) = x^2 – 2x + 2$
b. $ f(x) = \frac {ax- b}{cx -d}$
c. $ f(x) = \frac {1}{\sqrt {x-6}}$
d. $f(x) = x^2 – 5x + 6$
e. $f(x) = \frac {x-2}{5-x}$
I hope you like this article on how to find the range of a rational function
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