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.
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
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
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}
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}
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
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
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
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
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
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
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).
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
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).
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
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,
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
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,
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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