{"id":5425,"date":"2026-02-15T21:46:06","date_gmt":"2026-02-15T16:16:06","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=5425"},"modified":"2026-02-15T21:46:17","modified_gmt":"2026-02-15T16:16:17","slug":"find-the-range-of-a-function-algebraically","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/find-the-range-of-a-function-algebraically\/","title":{"rendered":"How to find the range of a function algebraically"},"content":{"rendered":"\n
\"how<\/figure>\n\n\n\n

For a function
f:A ->B<\/p>\n\n\n\n

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<\/p>\n\n\n\n

Now let’s see how to find the range of a function<\/a> algebraically i.e without plotting the graph<\/p>\n\n\n\n

How to find the range of a function algebraically<\/strong><\/p>\n\n\n\n

General Method is explained below. This is called the inverse function technique<\/p>\n\n\n\n

(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<\/p>\n\n\n\n

This is basically how to find the range of a function without graphing<\/p>\n\n\n\n

Let’s see fee examples with various types of functions<\/p>\n\n\n\n

How to find the range of a rational function<\/strong><\/h2>\n\n\n\n

a. $f(x) = \\frac {x-3}{x-1}$<\/p>\n\n\n\n

First, let’s see the domain of the function<\/a><\/p>\n\n\n\n

We can see that function is defined for all values of  x except 1<\/p>\n\n\n\n

So Domain is $R -{1}$<\/p>\n\n\n\n

Now let’s find the range using the inverse function method<\/p>\n\n\n\n

$y = \\frac {x-3}{x-1}$<\/p>\n\n\n\n

$y(x-1) = x-3$<\/p>\n\n\n\n

$xy -y = x-3$<\/p>\n\n\n\n

$3-y = x -xy$<\/p>\n\n\n\n

$x= \\frac {3-y}{1 -y}$<\/p>\n\n\n\n

It is very clear that x assumes real values for all y except y=1, So Range is<\/p>\n\n\n\n

$R – {1}$<\/p>\n\n\n\n

b. $f(x) = \\frac {x^2 -16}{x-4}$<\/p>\n\n\n\n

First let’s see the domain<\/p>\n\n\n\n

We can see that function is defined for all values of  x except 4<\/p>\n\n\n\n

So Range is $R -{4}$<\/p>\n\n\n\n

Now $f(x) = \\frac {x^2 -16}{x-4}$<\/p>\n\n\n\n

$f(x) = \\frac {(x-4)(x+4)}{x-4}$<\/p>\n\n\n\n

$f(x) = x+4$<\/p>\n\n\n\n

or y =x+4<\/p>\n\n\n\n

x= y -4<\/p>\n\n\n\n

It is very clear that x assumes real values for all y<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

So,Range is $R – {8}$<\/p>\n\n\n\n

How to find the range of a quadratic function\/polynomial function<\/strong><\/h2>\n\n\n\n

A quadratic Function \/Polynomial function<\/a> is like $f(x) = ax^2 + bx +c $.
a. $f(x) =x^2 -1 $<\/p>\n\n\n\n

Clearly, it is defined for all values of x, Domain =R<\/p>\n\n\n\n

Now<\/p>\n\n\n\n

$y =x^2 -1$<\/p>\n\n\n\n

$x^2 = y+1$<\/p>\n\n\n\n

$x = \\pm \\sqrt {y+1}$<\/p>\n\n\n\n

For x to be real , $y +1 \\geq 0$ or $ y \\geq -1$<\/p>\n\n\n\n

So $Range = [ -1 , \\infty )$<\/p>\n\n\n\n

b.  $f(x) =-2x^2 -1 $<\/p>\n\n\n\n

Clearly, it is defined for all values of x, Domain =R<\/p>\n\n\n\n

Now<\/p>\n\n\n\n

$y =-2x^2 -1$<\/p>\n\n\n\n

$y +1 =-2x^2$<\/p>\n\n\n\n

$ x^2 =\\frac {-1-y}{2}$<\/p>\n\n\n\n

$x = \\pm \\sqrt {\\frac  {-1-y}{2}}$<\/p>\n\n\n\n

For x to be real ,  $\\frac  {-1-y}{2} \\geq  0 $<\/p>\n\n\n\n

or $ -1 -y \\geq 0$<\/p>\n\n\n\n

$ -1 \\geq y$<\/p>\n\n\n\n

$ y \\leq -1$<\/p>\n\n\n\n

So $Range = (-\\infty\u00a0, -1]$<\/p>\n\n\n\n

How to find the range of a modulus function<\/strong><\/h2>\n\n\n\n

A modulus Function<\/a> is like $f(x) = |x-1|$
a. $f(x) = 1 – |x-3|$<\/p>\n\n\n\n

Clearly, it is defined for all values of x, Domain =R<\/p>\n\n\n\n

Now<\/p>\n\n\n\n

$y= 1 -|x-3|$<\/p>\n\n\n\n

$|x-3| = 1- y$<\/p>\n\n\n\n

Clearly for real values of x, $1-y \\geq 0$<\/p>\n\n\n\n

or<\/p>\n\n\n\n

$ 1 \\geq y$<\/p>\n\n\n\n

So Range is $(-\\infty , 1]$<\/p>\n\n\n\n

Similarly, we find the range of many functions algebraically i.e without plotting the graph. (how to find range of a function without graphing)<\/p>\n\n\n\n

Some Practice Questions<\/strong><\/p>\n\n\n\n

a. $ f(x) = x^2 – 2x + 2$<\/p>\n\n\n\n

b. $ f(x) = \\frac {ax- b}{cx -d}$<\/p>\n\n\n\n

c. $ f(x) = \\frac {1}{\\sqrt {x-6}}$<\/p>\n\n\n\n

d. $f(x) = x^2 – 5x + 6$<\/p>\n\n\n\n

e. $f(x) = \\frac {x-2}{5-x}$<\/p>\n\n\n\n

I hope you like this article on how to find the range of a rational function<\/p>\n\n\n\n

Related Articles<\/strong><\/p>\n\n\n\n

Class 11 Maths Formula book<\/a>
Important Solved Functions problems for JEE Maths<\/a>
What is Functions in Maths<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

For a functionf:A ->B Set A is called the domain of the function fSet B is called the co-domain of the functionThe 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 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a functionf:A ->B Set A is called the domain of the function fSet B is called the co-domain of the functionThe 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…","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/5425","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/comments?post=5425"}],"version-history":[{"count":3,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/5425\/revisions"}],"predecessor-version":[{"id":9915,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/5425\/revisions\/9915"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=5425"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=5425"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=5425"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}