For a function

f:A ->B

Set A is called the domain of the function f

Set B is the called the co-domain 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

Now lets 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 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 range of a function without graphing

Lets see fee examples with various type of functions

**How to find the range of a rational function**

a. $f(x) = \frac {x-3}{x-1}$

First lets 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 lets 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 lets 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 function algebraically i.e without plotting the graph.

**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. $f(x) = \frac {x-2}{5-x}$

**Related Articles**

Important Solved Functions problems for JEE Maths

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