How to Find the Range of a Function Algebraically [15 Ways]

There are different ways to Find the Range of a Function Algebraically. But before that, we take a short overview of the Range of a Function.

In the first chapter What is a Function? we have learned that a function is expressed as

y=f(x),

where x is the input and y is the output.

For every input x (where the function f(x) is defined) there is a unique output.

The set of all outputs of a function is the Range of a Function.

How to Find the Range of a Function Algebraically

The Range of a Function is the set of all y values or outputs i.e., the set of all f(x) when it is defined.

We suggest you read this article “9 Ways to Find the Domain of a Function Algebraically” first. This will help you to understand the concepts of finding the Range of a Function better.

In this article, you will learn

  1. 5 Steps to Find the Range of a Function,

and in the end you will be able to

  1. Find the Range of 10 different types of functions

Steps to Find the Range of a Function

Suppose we have to find the range of the function f(x)=x+2.

We can find the range of a function by using the following steps:

#1. First label the function as y=f(x)

y=x+2

#2. Express x as a function of y

Here x=y-2

#3. Find all possible values of y for which f(y) is defined

See that x=y-2 is defined for all real values of y.

#4. Element values of y by looking at the initial function f(x)

Our initial function y=x+2 is defined for all real values of x i.e., x\epsilon \mathbb{R}.

So here we do not need to eliminate any value of y i.e., y\epsilon \mathbb{R}.

#5. Write the Range of the function f(x)

Therefore the Range of the function y=x+2 is {y\epsilon \mathbb{R}}.

Maybe you are getting confused and don’t understand all the steps now.

But believe me, you will get a clear concept in the next examples.


How to Find the Range of a Function Algebraically

There are different types of functions. Here you will learn 10 ways to find the range for each type of function.

#1. Find the range of a Rational function

Example 1: Find the range

f(x)=\frac{x-2}{3-x},x\neq3

Solution:

Step 1: First we equate the function with y

y=\frac{x-2}{3-x}

Step 2: Then express x as a function of y

y=\frac{x-2}{3-x}

or, y(3-x)=x-2

or, 3y-xy=x-2

or, x+xy=3y+2

or, x(1+y)=3y+2

or, x=\frac{3y+2}{y+1}

Step 3: Find possible values of y for which x=f(y) is defined

x=\frac{3y+2}{y+1} is defined when y+1 can not be equal to 0,

i.e., y+1\neq0

i.e., y\neq-1

i.e., y\epsilon \mathbb{R}-{-1}

Step 4: Eliminate the values of y

See that f(x)=\frac{x-2}{3-x} is defined on \mathbb{R}-{3} and we do not need to eliminate any value of y from y\epsilon \mathbb{R}-{-1}.

Step 5: Write the Range

\therefore the range of f(x)=\frac{x-2}{3-x} is {x\epsilon \mathbb{R}:x\neq-1}.


Example 2: Find the range

f(x)=\frac{3}{2-x^{2}}

Solution:

Step 1:

y=\frac{3}{2-x^{2}}

Step 2:

y=\frac{3}{2-x^{2}}

or, 2y-xy^{2}=3

or, 2y-3=x^{2y}

or, x^{2}=\frac{2y-3}{y}

Step 3:

The function x^{2}=\frac{2y-3}{y} is defined when y\neq 0 …(1)

Also since x^{2}\geq 0,

therefore

\frac{2y-3}{y}\geq 0

or, \frac{2y-3}{y}\times {\color{Magenta} \frac{y}{y}}\geq 0

or, \frac{y(2y-3)}{y^{2}}\geq 0

or, y(2y-3)\geq 0 (\because y^{2}\geq 0)

or, (y-0){\color{Magenta} 2}(y-\frac{3}{{\color{Magenta} 2}})

or, (y-0)(y-\frac{3}{2})\geq 0

Next we find the values of y for which (y-0)(y-\frac{3}{2})\geq 0 i.e., y(2y-3)\geq 0 is satisfied.

Now see the table:

Value of ySign of (y-0)Sign of (2y-3)Sign of y(2y-3)y(2y-3)\geq 0 satisfied or not
y=-1<0
i.e.,
y\epsilon (-\infty,0)
-ve-ve+ve
i.e., >0
y=00-ve=0
y=1
i.e.,
y\epsilon (0,\frac{3}{2})
+ve-ve-ve
i.e., <0
y=\frac{3}{2}+ve0=0
y=2>\frac{3}{2}
i.e.,
y\epsilon (\frac{3}{2},\infty)
+ve+ve+ve
i.e., >0

Therefore from the above table and using (1) we get,

y\epsilon (-\infty,0)\cup [\frac{3}{2},\infty) (\because y\neq 0)

Step 4:

y=\frac{3}{2-x^{2}} is not a square function,

\therefore we do not need to eliminate any value of y except 0 because if y be zero then the function y=\frac{3}{2-x^{2}} will be undefined.

Step 5:

Therefore the range of the function f(x)=\frac{3}{2-x^{2}} is

(-\infty,0)\cup [\frac{3}{2},\infty).


Example 3: Find the range of a rational equation using inverse

f(x)=\frac{2x-1}{x+4}

Solution:

Video Source: YouTube | Video by: Brian McLogan (Duration: 6 minutes 38 seconds)

#2. Find the range of a function with square root

Example 4: Find the range

f(x)=\sqrt{4-x^{2}}

Solution:

Step 1: First we equate the function with y

y=\sqrt{4-x^{2}}

Step 2: Then express x as a function of y

y=\sqrt{4-x^{2}}

or, y^{2}=4-x^{2}

or, x^{2}=4-y^{2}

Step 3: Find possible values of y for which x=f(y) is defined

Since x^{2}\geq 0,

\therefore 4-y^{2}\geq 0

or, (2-y)(2+y)\geq 0

or, (y-2)(y+2)\leq 0

Now we find possible values for which (y-2)(y+2)\leq 0

Value of ySign of (y-2)Sign of (y+2)Sign of (y-2)(y+2)(y-2)(y+2)\leq 0 is satisfied or not
y=-3<-2
i.e., y\epsilon (-\infty,-2)
-ve-ve+ve
i.e., >0
y=-2-ve0=0
y=0
i.e., -2<y<2
i.e., y\epsilon (-2,2)
-ve+ve-ve
i.e., <0
y=20+ve=0
y=3>2
i.e., y\epsilon (2,\infty)
+ve+ve+ve
i.e., >0

i.e., y=-2, y\epsilon (-2,2) and y=2

i.e., y\epsilon [-2,2]

Step 4: Eliminate the values of y

As y=\sqrt{4-x^{2}}, a square root function,

so y can not take any negative value i.e., y\geq 0

Therefore y\epsilon [0,2].

Step 5: Write the range

The range of the function f(x)=\sqrt{4-x^{2}} is [0,2] in interval notation.

We can also write the range of the function f(x)=\sqrt{4-x^{2}} as R(f)={x\epsilon \mathbb{R}:0\leq y \leq 2}


Example 5: Find the range of a function f(x) =\sqrt{x^{2}-4}.

Solution:

Step 1: First we equate the function with y

y=\sqrt{x^{2}-4}

Step 2: Then express x as a function of y

y=\sqrt{x^{2}-4}

or, y^{2}=x^{2}-4

or, x^{2}=y^{2}+4

Step 3: Find possible values of y for which x=f(y) is defined

Since x^{2}\geq 0,

therefore y^{2}+4\geq 0

i.e., y^{2}\geq -4

i.e., y\geq \sqrt{-4}

i.e, y\geq i\sqrt{2}, a complex number

\therefore y^{2}+4\geq 0 for all y\epsilon \mathbb{R}

Step 4: Eliminate the values of y

Since y=\sqrt{x^{2}-4} is a square root function,

therefore y can not take any negative value i.e., y\geq 0

Step 5: Write the Range

The range of f(x) =\sqrt{x^{2}-4} is (0,\infty).


Example 6: Find the range for the square root function

f(x)=3-\sqrt{x}

Solution:

Video Source: YouTube | Video by: Brian McLogan (duration: 3 minutes 5 seconds)

#3. Find the range of a function with a square root in the denominator

Example 7: Find the range

f(x)=\frac{1}{\sqrt{x-3}}

Solution:

Step 1:

y=\frac{1}{\sqrt{x-3}}

Step 2:

y=\frac{1}{\sqrt{x-3}}

or, y=\frac{1}{\sqrt{x-3}}

or, y^{2}=\frac{1}{x-3}

or, y^{2}(x-3)=1

or, xy^{2}-3y^{2}=1

or, xy^{2}=1+3y^{2}

or, x=\frac{1+3y^{2}}{y^{2}}

Step 3:

For x=\frac{1+3y^{2}}{y^{2}} to be defined,

y^{2}\neq 0

i.e., y\neq 0

Step 4:

As f(x)=\frac{1}{\sqrt{x-3}}, so y can not be negative (-ve).

Step 5:

The range of f(x)=\frac{1}{\sqrt{x-3}} is (0,\infty).


Example 8: Find the range

f(x)=\frac{1}{\sqrt{4-x^{2}}}

Solution:

Step 1:

y=\frac{1}{\sqrt{4-x^{2}}}

Step 2:

y=\frac{1}{\sqrt{4-x^{2}}}

or, y=\frac{1}{\sqrt{4-x^{2}}}

or, y^{2}=\frac{1}{4-x^{2}}

or, 4y^{2}-x^{2}y^{2}=1

or, x^{2}y^{2}=4y^{2}-1

or, x^{2}=\frac{4y^{2}-1}{y^{2}}

Step 3:

For x^{2}=\frac{4y^{2}-1}{y^{2}} to be defined, y can not be equal to zero

i.e., y\neq 0

Also since x^{2}\geq 0,

\therefore \frac{4y^{2}-1}{y^{2}}\geq 0

or, 4y^{2}-1\geq 0 (\because y^{2}\geq 0)

or, (2y-1)(2y+1)\geq 0

or, 4(y-\frac{1}{2})(y+\frac{1}{2})\geq 0

Value of ySign of (2y-1)Sign of (2y+1)Sign of (2y-1)(2y+1)(2y-1)(2y+1)\geq 0 is satisfied or not
y=-1<-\frac{1}{2}
i.e., y\epsilon \left ( -\infty,-\frac{1}{2} \right )
-ve-ve+ve
i.e., >0
y=-\frac{1}{2}-ve00
y=0
i.e., y\epsilon \left (-\frac{1}{2},\frac{1}{2} \right )
-ve+ve-ve
i.e., <0
y=\frac{1}{2}0+ve+ve
i.e., >0
y=1>\frac{1}{2}
i.e., y\epsilon \left (\frac{1}{2},\infty \right )
+ve+ve+ve
i.e., >0

The above table implies that

y\epsilon \left ( -\infty,-\frac{1}{2} \right )\cup \left (\frac{1}{2},\infty \right ) …..(1)

Step 4:

Since y=\frac{1}{\sqrt{4-x^{2}}} is a square root function,

therefore y can not be negative (-ve).

i.e., y\geq 0 …..(2)

Now from (1) and (2), we get

y\epsilon ( \frac{1}{2},\infty )

Step 5:

Therefore the range of the function f(x)=\frac{1}{\sqrt{4-x^{2}}} is [ \frac{1}{2},\infty )


Example 9: Find the range of the function

f(x)=\sqrt{\frac{(x-3)(x+2)}{x-1}}

Solution:

Video Source: YouTube | Video by: Anil Kumar (Duration: 10 minutes 17 seconds)

#4. Find the range of modulus function or absolute value function

Example 10: Find the range of the absolute value function

f(x)=\left | x \right |

Solution:

We can find the range of the absolute value function f(x)=\left | x \right | on a graph.

If we draw the graph then we get

Find the range of modulus function or absolute value function

Here you can see that the y value starts at y=0 and extended to infinity.

\therefore the range of the absolute value function f(x)=\left | x \right | is [0,\infty).


Example 11: Find the range of the absolute value function

f(x)=-\left | x-1 \right |

Solution:

The graph of f(x)=-\left | x-1 \right | is

Find the range of modulus function or absolute value function

From the graph, it is clear that the y value starts from y=0 and extended to -\infty.

Therefore the range of f(x)=-\left | x-1 \right | is (-\infty,0].


Shortcut Trick:

  1. If the sign before modulus is positive (+ve) i.e., of the form +\left | x-a \right |, then the range will be [a,\infty),
  2. If the sign before modulus is negative (-ve) i.e., of the form -\left | x-a \right |, then the range will be (-\infty,a].

We can also find the range of the absolute value functions f(x)=\left | x \right | and f(x)=-\left | x-1 \right | using the above short cut trick:

The function f(x)=\left | x \right | can be written as f(x)=+\left | x-0 \right |

Now using trick 1 we can say, the range of f(x)=\left | x \right | is [0,\infty)

Also using trick 2 we can say, the range of f(x)=-\left | x-1 \right | is (-\infty,0].


Example 12: Find the range of the following absolute value functions

  1. f(x)=\left | x \right |+6,
  2. f(x)=\left | x+4 \right |

Solution:

Video Source: YouTube | Video by: Brian Nelson (Duration: 6 minutes 28 seconds)

#5. Find the range of a Step function

Example 13: Find the range of the step function f(x)=[x],x\epsilon \mathbb{R}.

Solution:

The step function f(x)=[x],x\epsilon \mathbb{R} is expressed as

f(x)=0, 0\leq x<1

=1, 1\leq x<2

=2,2\leq x<3

………

=-1,-1\leq x<0

=-2,-2\leq x<-1

………

You can verify this result from the graph of f(x)=[x],x\epsilon \mathbb{R}

Find the range of a Step function

i.e., y\epsilon {…,-2,-1,0,1,2,…}

i.e., y\epsilon \mathbb{Z}, the set of all integers.

\therefore the range of the step function f(x)=[x],x\epsilon \mathbb{R} is \mathbb{Z}, the set of all integers.


Example 14: Find the range of the step function f(x)=[x-3],x\epsilon \mathbb{R}.

Solution:

By using the definition of step function, we can express f(x)=[x-3],x\epsilon \mathbb{R} as

f(x)=1,3\leq x<4

=2,4\leq x<5

=3,5\leq x<6

………

=0,2\leq x<3

=-1,1\leq x<2

=-2, 0\leq x<1

=-3, -1\leq x<0

………

You can verify this result from the graph of f(x)=[x-3],x\epsilon \mathbb{R}

Find the range of a Step function

i.e., y\epsilon {…,-3,-2,-1,0,1,2,3,…}

i.e., y\epsilon \mathbb{Z}, the set of all integers.

\therefore the range of the step function f(x)=[x-3],x\epsilon \mathbb{R} is \mathbb{Z}, the set of all integers.


Example 15: Find the range of the step function f(x)=\left [ \frac{1}{4x} \right ],x\epsilon \mathbb{R}.

Solution:

Video Source: YouTube | Video by: Jessica Tentinger (Duration: 3 minutes 32 seconds)

#6. Find the range of an Exponential function

Example 16: Find the range of the exponential function f(x)=2^{x}.

Solution:

The graph of the function f(x)=2^{x} is

Find the range of an Exponential function

Here y=0 is an asymptote of f(x)=2^{x} i.e., the graph is going very close and close to the y=0 straight line but it will never touch y=0.

Also, you can see on the graph that the function is extended to +\infty.

So we can say y>0.

\therefore the range of the exponential function f(x)=2^{x} is (0,\infty).


Example 17: Find the range of the exponential function

f(x)=-3^{x+1}+2.

Solution:

The graph of the exponential function f(x)=-3^{x+1}+2 is

Find the range of an Exponential function

From the graph of f(x)=-3^{x+1}+2 you can see that y=2 is an asymptote of f(x)=-3^{x+1}+2 i.e., on the graph f(x)=-3^{x+1}+2 is going very close and close to y=2 towards -ve x-axis but it will never touch the straight line y=2 and extended to -\infty towards +ve x-axis.

i.e., y<2

\therefore the range of the exponential function f(x)=-3^{x+1}+2 is (-\infty,2).

There is a shortcut trick to find the range of any exponential function. This trick will help you find the range of any exponential function in just 2 seconds.


Shortcut trick:

Let f(x)=a\times b^{x-h}+k be an exponential function.

Then

  1. If a>0, then R(f)=(k,\infty),
  2. If a<0, then R(f)=(-\infty,k).

Now we try to find the range of the exponential functions f(x)=2^{x} and f(x)=-3^{x+1}+2 with the above shortcut trick:

We can write f(x)=2^{x} as f(x)=1\times 2^{x}+0, 1>0 and comparing this result with trick 1 we directly say

The range of f(x)=2^{x} is (0,\infty).

Also f(x)=-3^{x+1}+2 can be written as f(x)=-1\times 3^{x+1}+2, -1<0 and comparing with trick 2 we get

The range of f(x)=-3^{x+1}+2 is (-\infty,2).


Example 18: Find the range of the exponential functions given below

f(x)=-2^{x+1}+3

Solution:

Video Source: YouTube | Video by: Daytona State College Instructional Resources (Duration: 6 minutes 25 seconds)

#7. Find the range of a Logarithmic function

The range of any logarithmic function is (-\infty,\infty).

We can verify this fact from the graph.

f(x)=\log_{2}x^{3} is a logarithmic function and the graph of this function is

Find the range of a Logarithmic function

Here you can see that the y value starts from -\infty and extended to +\infty,

i.e., the range of f(x)=\log_{2}x^{3} is (-\infty,\infty).


Example 19: Find the range of the logarithmic function

f(x)=\log_{2}(x+4)+3

Solution:

Video Source: YouTube | Video by: Daytona State College Instructional Resources (Duration: 5 minutes 22 seconds)

#8. Find the range of a function relation of ordered pairs

A relation is the set of ordered pairs i.e., the set of (x,y) where the set of all x values is called the domain and the set of all y values is called the range of the relation.

In the previous chapter, we have learned how to find the domain of a function using relation.

Now we learn how to find the range of a function using relation.

For that we have to remember 2 rules which are given below:

Rules:

  1. Before finding the range of a function first we check the given relation (i.e., the set of ordered pairs) is a function or not
  2. Find all the y values and form a set. This set is the range of the relation.

Now see the examples given below to understand this concept:


Example 20: Find the range of the relation

{(1,3), (5,9), (8,23), (12,14)}

Solution:

In the relation {(1,3), (5,9), (8,23), (12,14)}, the set of x coordinates is {1, 5, 8, 12} and the set of y coordinates is {3, 9, 14, 23}.

If we draw the diagram of the given relation it will look like this

How to Find the Range of a Function Relation, How to Find the Range of a Function Ordered Pairs

Here we can clearly see that each element of the set {1, 5, 8, 12} is related to a unique element of the set {3, 9, 14, 23}.

Therefore the given relation is a Function.

Also, we know that the range of a function relation is the set of y coordinates.

Therefore the range of the relation {(1,3), (5,9), (8,23), (12,14)} is the set {3, 9, 14, 23}.


Example 21: Find the range of the set of ordered pairs

{(5,2), (7,6), (9,4), (9,13), (12,19)}.

Solution:

The diagram of the given relation is

How to Find the Range of a Function Relation, How to Find the Range of a Function Ordered Pairs

Here we can see that element 9 is related to two different elements and they are 4 and 13 i.e., 9 is not related to a unique element and this goes against the definition of the function.

Therefore the relation {(5,2), (7,6), (9,4), (9,13), (12,19)} is not a Function.


Example 22: Determine the range of the relation described by the table

xy
-13
3-2
32
48
6-1

Solution:

Video Source: YouTube | Video by: Khan Academy (Duration: 2 minutes 42 seconds)

#9. Find the range of a Discrete function

A Discrete Function is a collection of some points on the Cartesian plane and the range of a discrete function is the set of y-coordinates of the points.

Example 23: How do you find the range of the discrete function from the graph

How to find the Range of a Discrete Function

Solution:

From the graph, we can see that there are five points on the discrete function and they are A (2,2), B (4,4), C (6,6), D (8,8), and E (10,10).

How to find the Range of a Discrete Function

The set of the y-coordinates of the points A, B, C, D, and E is {2,4,6, 8, 10}.

\therefore the range of the discrete function is {2,4,6,8,10}.


Example 24: Find the range of the discrete function from the graph

How to find the Range of a Discrete Function

Solution:

The discrete function is made of the five points A (-3,2), B (-2,4), C (2,3), D (3,1), and E (5,5).

How to find the Range of a Discrete Function

The set of the y coordinates of the discrete function is {2,4,3,1,5} = {1,2,3,4,5}.

\therefore the range of the discrete function is {1,2,3,4,5}.

Video Source: YouTube | Video by: Jillian Tomsche (Duration: 9 minutes 55 seconds)

#10. Find the range of a trigonometric function

Trigonometric FunctionExpresionRange
Sine function\sin x[-1,1]
Cosine function\cos x[-1,1]
Tangent function\tan x(-\infty,+\infty)
CSC function
(Cosecant function)
\csc x(-\infty,-1]\cup[1,+\infty)
Secant function\sec x(-\infty,-1]\cup[1,+\infty)
Cotangent function\sec x(-\infty,+\infty)

#11. Find the range of an inverse trigonometric function

Inverse trigonometric functionExpressionRange
Arc Sine function /
Inverse Sine function
\arcsin x
or, \sin^{-1}x
[-\frac{\pi}{2},+\frac{\pi}{2}]
Arc Cosine function /
Inverse Cosine function
\arccos x
or, \cos^{-1}x
[0,\pi]
Arc Tangent function /
Inverse Tangent function
\arctan x
or, \tan^{-1}x
(-\frac{\pi}{2},+\frac{\pi}{2})
Arc CSC function /
Inverse CSC function
\textrm{arccsc}x
or, \csc^{-1}x
[-\frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]
Arc Secant function /
Inverse Secant function
\textrm{arcsec}x
or, \sec^{-1}x
[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]
Arc Cotangent function /
Inverse Cotangent function
\textrm{arccot}x
or, \cot^{-1}x
(0,\pi)

#12. Find the range of a hyperbolic function

Hyperbolic functionExpressionRange
Hyperbolic Sine function\sinh x=\frac{e^{x}-e^{-x}}{2}(-\infty,+\infty)
Hyperbolic Cosine function\cosh x=\frac{e^{x}+e^{-x}}{2}[1,\infty)
Hyperbolic Tangent function\tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}(-1,+1)
Hyperbolic CSC functioncsch x=\frac{2}{e^{x}-e^{-x}}(-\infty,0)\cup(0,\infty)
Hyperbolic Secant functionsech x=\frac{2}{e^{x}+e^{-x}}(0,1)
Hyperbolic Cotangent function\tanh x=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}(-\infty,-1)\cup(1,\infty)

#13. Find the range of an inverse hyperbolic function

Inverse hyperbolic functionExpressionRange
Inverse hyperbolic sine function\sinh^{-1}x=\ln(x+\sqrt{x^{2}+1})(-\infty,\infty)
Inverse hyperbolic cosine function\cosh^{-1}x=\ln(x+\sqrt{x^{2}-1})[0,\infty)
Inverse hyperbolic tangent function\tanh^{-1}x=\frac{1}{2}\ln\left (\frac{1+x}{1-x}\right )(-\infty,\infty)
Inverse hyperbolic CSC functioncsch^{-1}x=\ln \left ( \frac{1+\sqrt{1+x^{2}}}{x} \right )(-\infty,0)\cup(0,\infty)
Inverse hyperbolic Secant functionsech^{-1}x=\ln \left ( \frac{1+\sqrt{1-x^{2}}}{x} \right )[0,\infty)
Inverse hyperbolic Cotangent functioncoth^{-1}x=\frac{1}{2}\ln\left (\frac{x+1}{x-1}\right )(-\infty,0)\cup(0,\infty)

#14. Find the range of a piecewise function

Example 25: Find the range of the piecewise function

Piecewise function

Solution:

The piecewise function consists of two function:

  1. f(x)=x-3 when x\leq -1,
  2. f(x)=x+1 when x>1.

If we plot these two functions on the graph then we get,

Find the range of a piecewise function

This is the graph of the piecewise function.

From the graph, we can see that

  1. the range of the function f(x)=x-3 is (-\infty,-2] when x\leq -1,
  2. the range of the function f(x)=x+1 is (2,\infty) when x>1,

Therefore from the above results we can say that

The range of the piecewise function f(x) is

(-\infty,-2]\cup (2,\infty).


Example 26: Find the range of a piecewise function given below

Piecewise function

Solution:

If you notice the piecewise function then you can see there are functions:

  1. f(x)=x defined when x\leq -1,
  2. f(x)=2 defined when -1<x<1),
  3. f(x)=\sqrt{x} defined when x\geq 1.

Now if we draw the graph of these three functions we get,

Find the range of a piecewise function

This is the graph of the piecewise function.

Here you can see that

The function f(x)=x starts y=-1 and extended to -\infty when x\leq -1.

So the range of the function f(x)=x,x\leq -1 is (-\infty,-1]……..(1)

The functional value of the function f(x)=2, -1<x<1 is 2.

The range of the function f(x)=x is {2}……..(2)

The function f(x)=\sqrt{x} starts at y=1 and extended to \infty when x\geq 1.

The range of the function f(x)=\sqrt{x} is [1,\infty) when x\geq 1……..(3)

From (1), (2), and (3), we get,

the range of the piecewise function is

(-\infty,-1]\cup {2}\cup [1,\infty)

= (-\infty,-1]\cup [1,\infty)

Video Source: YouTube | Video by: patrickJMT (Duration: 4 minutes 55 seconds)

#15. Find the range of a composite function

Example 27: Let f(x)=2x-6 and g(x)=\sqrt{x} be two functions.

Find the range of the following composite functions:

(a) f\circ g(x)

(b) g\circ f(x)

Solution of (a)

First we need to find the function g\circ f(x).

We know that,

f\circ g(x)

=f(g(x))

=f(\sqrt{x}) (\because g(x)=\sqrt{x})

=2\sqrt{x}-6

Now see that 2\sqrt{x}-6 is a function with a square root and at the beginning of this article, we already learned how to find the range of a function with a square root.

Following these steps, we can get,

the range of the composite function f of g is

R(f\circ g)=[-6,\infty).

Solution of (b):

g\circ f(x)

=g(f(x))

=g(2x-6) (\because f(x)=2x-6)

=\sqrt{2x-6}, a function with a square root

Using the previous method we get,

the range of the composite function g\circ f(x) is

R(g\circ f(x))=[0,\infty)


Example 28: Let f(x)=3x-12 and g(x)=\sqrt{x} be two functions.

Find the range of the following composite functions

  1. f\circ g(x),
  2. g\circ f(x)

Solution:

Video Source: YouTube | Video by: 
Anil Kumar (Duration: 3 minutes 35 seconds)

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