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 [15 Ways] How to Find the Range of a Function Algebraically](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/How-to-Find-the-Range-of-a-Function-Algebraically-2.png?resize=482%2C271&ssl=1)
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
and in the end you will be able to
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 y | Sign 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=0 | 0 | -ve | =0 | ✅ |
y=1 i.e., y\epsilon (0,\frac{3}{2}) | +ve | -ve | -ve i.e., <0 | ❌ |
y=\frac{3}{2} | +ve | 0 | =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:
#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 y | Sign 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 | -ve | 0 | =0 | ✅ |
y=0 i.e., -2<y<2 i.e., y\epsilon (-2,2) | -ve | +ve | -ve i.e., <0 | ✅ |
y=2 | 0 | +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:
#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 0or, 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 y | Sign 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} | -ve | 0 | 0 | ✅ |
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:
#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
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of modulus function or absolute value function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/Find-the-range-of-modulus-function-absolute-value-function-1.png?resize=700%2C375&ssl=1)
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
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of modulus function or absolute value function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/Find-the-range-of-modulus-function-absolute-value-function-2.png?resize=700%2C375&ssl=1)
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:
- If the sign before modulus is positive (+ve) i.e., of the form +\left | x-a \right |, then the range will be [a,\infty),
- 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
- f(x)=\left | x \right |+6,
- f(x)=\left | x+4 \right |
Solution:
#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}
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of a Step function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/Find-the-range-of-a-Step-function-1.png?resize=700%2C375&ssl=1)
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}
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of a Step function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/Find-the-range-of-a-Step-function-2.png?resize=700%2C375&ssl=1)
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:
#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
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of an Exponential function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/Find-the-range-of-an-Exponential-function-1.png?resize=700%2C375&ssl=1)
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
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of an Exponential function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/Find-the-range-of-an-Exponential-function-2.png?resize=700%2C375&ssl=1)
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
- If a>0, then R(f)=(k,\infty),
- 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:
#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
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of a Logarithmic function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/Find-the-range-of-a-Logarithmic-function-1.png?resize=700%2C375&ssl=1)
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:
#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:
- 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
- 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 Algebraically [15 Ways] How to Find the Range of a Function Relation, How to Find the Range of a Function Ordered Pairs](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/How-to-Find-the-range-of-a-function-relation-of-ordered-pairs-1.png?resize=560%2C315&ssl=1)
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 Algebraically [15 Ways] How to Find the Range of a Function Relation, How to Find the Range of a Function Ordered Pairs](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/How-to-Find-the-range-of-a-function-relation-of-ordered-pairs-2.png?resize=560%2C315&ssl=1)
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
x | y |
---|---|
-1 | 3 |
3 | -2 |
3 | 2 |
4 | 8 |
6 | -1 |
Solution:
#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 Function Algebraically [15 Ways] How to find the Range of a Discrete Function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/how-to-find-the-range-of-a-discrete-function-1.png?resize=700%2C375&ssl=1)
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 Function Algebraically [15 Ways] How to find the Range of a Discrete Function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/how-to-find-the-range-of-a-discrete-function-2.png?resize=700%2C375&ssl=1)
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 Function Algebraically [15 Ways] How to find the Range of a Discrete Function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/how-to-find-the-range-of-a-discrete-function-3.png?resize=700%2C375&ssl=1)
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 Function Algebraically [15 Ways] How to find the Range of a Discrete Function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/how-to-find-the-range-of-a-discrete-function-4.png?resize=700%2C375&ssl=1)
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}.
#10. Find the range of a trigonometric function
Trigonometric Function | Expresion | Range |
---|---|---|
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 function | Expression | Range |
---|---|---|
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 function | Expression | Range |
---|---|---|
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 function | csch x=\frac{2}{e^{x}-e^{-x}} | (-\infty,0)\cup(0,\infty) |
Hyperbolic Secant function | sech 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 function | Expression | Range |
---|---|---|
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 function | csch^{-1}x=\ln \left ( \frac{1+\sqrt{1+x^{2}}}{x} \right ) | (-\infty,0)\cup(0,\infty) |
Inverse hyperbolic Secant function | sech^{-1}x=\ln \left ( \frac{1+\sqrt{1-x^{2}}}{x} \right ) | [0,\infty) |
Inverse hyperbolic Cotangent function | coth^{-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
![How to Find the Range of a Function Algebraically [15 Ways] Piecewise function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/piecewise-function-1.png?resize=373%2C105&ssl=1)
Solution:
The piecewise function consists of two function:
- f(x)=x-3 when x\leq -1,
- f(x)=x+1 when x>1.
If we plot these two functions on the graph then we get,
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of a piecewise function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/find-the-range-of-a-piecewise-function-1.png?resize=700%2C375&ssl=1)
This is the graph of the piecewise function.
From the graph, we can see that
- the range of the function f(x)=x-3 is (-\infty,-2] when x\leq -1,
- 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
![How to Find the Range of a Function Algebraically [15 Ways] Piecewise function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/piecewise-function-2.png?resize=431%2C149&ssl=1)
Solution:
If you notice the piecewise function then you can see there are functions:
- f(x)=x defined when x\leq -1,
- f(x)=2 defined when -1<x<1),
- f(x)=\sqrt{x} defined when x\geq 1.
Now if we draw the graph of these three functions we get,
![How to Find the Range of a Function Algebraically [15 Ways] Find the range of a piecewise function](https://i0.wp.com/mathculus.com/wp-content/uploads/2020/10/find-the-range-of-a-piecewise-function-2.png?resize=700%2C375&ssl=1)
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)
#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(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
- f\circ g(x),
- g\circ f(x)
Solution:
Anil Kumar (Duration: 3 minutes 35 seconds)
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Very nicely explained just loved your hard work! Thanks for helping mate. Cheers