How To Find The Range Of A Function Algebraically
There are different means to Find the Range of a Role Algebraically. But before that, we take a short overview of the Range of a Role.
In the outset chapter What is a Function? we have learned that a office is expressed as
y=f(10),
where x is the input and y is the output.
For every input x (where the part f(x) is divers) there is a unique output.
The prepare of all outputs of a function is the Range of a Function.
The Range of a Part is the ready of all y values or outputs i.e., the set of all f(x) when it is defined.
We suggest yous read this article "9 Ways to Discover the Domain of a Part Algebraically" first. This will help you lot to understand the concepts of finding the Range of a Function better.
In this article, you will acquire
- v Steps to Find the Range of a Function,
and in the finish you volition be able to
- Find the Range of 10 unlike types of functions
Table of Contents - What yous will larn
Steps to Observe the Range of a Function
Suppose we take to find the range of the part f(ten)=x+ii.
Nosotros tin find the range of a function past using the following steps:
#ane. First characterization the part as y=f(x)
y=x+2
#ii. Limited x every bit a function of y
Hither x=y-2
#3. Find all possible values of y for which f(y) is defined
Run into that x=y-2 is defined for all real values of y.
#four. Chemical element values of y by looking at the initial part f(x)
Our initial office y=x+2 is defined for all existent values of ten i.e., x\epsilon \mathbb{R}.
And so here we do non 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 part y=x+2 is {y\epsilon \mathbb{R}}.
Maybe you lot are getting confused and don't understand all the steps now.
Merely believe me, you lot will get a clear concept in the next examples.
How to Notice the Range of a Function Algebraically
There are different types of functions. Here you lot will larn 10 means to notice the range for each type of function.
#ane. Notice the range of a Rational function
Example one: Notice the range
f(ten)=\frac{x-ii}{3-x},x\neq3
Solution:
Footstep ane: Beginning we equate the function with y
y=\frac{ten-2}{3-10}
Stride 2: Then express ten as a office of y
y=\frac{x-two}{3-x}
or, y(three-10)=x-2
or, 3y-xy=x-ii
or, x+xy=3y+2
or, ten(one+y)=3y+2
or, x=\frac{3y+ii}{y+i}
Step 3: Find possible values of y for which 10=f(y) is defined
10=\frac{3y+2}{y+1} is defined when y+i can not exist equal to 0,
i.eastward., y+1\neq0
i.e., y\neq-1
i.eastward., y\epsilon \mathbb{R}-{-1}
Step 4: Eliminate the values of y
Encounter that f(x)=\frac{x-2}{iii-x} is defined on \mathbb{R}-{iii} and we practice not need to eliminate whatsoever value of y from y\epsilon \mathbb{R}-{-ane}.
Step 5: Write the Range
\therefore the range of f(10)=\frac{x-2}{3-x} is {x\epsilon \mathbb{R}:x\neq-ane}.
Example two: Notice the range
f(x)=\frac{3}{two-x^{2}}
Solution:
Step 1:
y=\frac{3}{2-x^{2}}
Step 2:
y=\frac{3}{two-x^{2}}
or, 2y-xy^{2}=3
or, 2y-3=x^{2y}
or, ten^{2}=\frac{2y-three}{y}
Stride 3:
The function x^{2}=\frac{2y-3}{y} is defined when y\neq 0 …(1)
As well since x^{2}\geq 0,
therefore
\frac{2y-3}{y}\geq 0
or, \frac{2y-three}{y}\times {\color{Magenta} \frac{y}{y}}\geq 0
or, \frac{y(2y-three)}{y^{2}}\geq 0
or, y(2y-iii)\geq 0 (\because y^{two}\geq 0)
or, (y-0){\color{Magenta} 2}(y-\frac{3}{{\color{Magenta} 2}})
or, (y-0)(y-\frac{iii}{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 run across the tabular array:
| Value of y | Sign of (y-0) | Sign of (2y-3) | Sign of y(2y-3) | y(2y-three)\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}{two} | +ve | 0 | =0 | ✅ |
| y=2>\frac{three}{two} i.due east., y\epsilon (\frac{3}{2},\infty) | +ve | +ve | +ve i.east., >0 | ✅ |
Therefore from the above table and using (i) we get,
y\epsilon (-\infty,0)\cup [\frac{iii}{2},\infty) (\because y\neq 0)
Footstep 4:
y=\frac{three}{2-x^{ii}} 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}{2-10^{2}} volition exist undefined.
Stride five:
Therefore the range of the function f(10)=\frac{3}{two-10^{2}} is
(-\infty,0)\cup [\frac{3}{2},\infty).
Instance three: Find the range of a rational equation using changed
f(x)=\frac{2x-1}{x+4}
Solution:
#2. Discover the range of a function with square root
Example 4: Find the range
f(x)=\sqrt{iv-x^{ii}}
Solution:
Footstep 1: Beginning we equate the office with y
y=\sqrt{4-10^{2}}
Step 2: Then express x as a function of y
y=\sqrt{4-10^{ii}}
or, y^{2}=4-x^{2}
or, ten^{ii}=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)(ii+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-ii) | Sign of (y+2) | Sign of (y-2)(y+2) | (y-ii)(y+ii)\leq 0 is satisfied or non |
|---|---|---|---|---|
| y=-3<-two i.e., y\epsilon (-\infty,-2) | -ve | -ve | +ve i.east., >0 | ❌ |
| y=-2 | -ve | 0 | =0 | ✅ |
| y=0 i.e., -2<y<2 i.due east., y\epsilon (-2,2) | -ve | +ve | -ve i.due east., <0 | ✅ |
| y=2 | 0 | +ve | =0 | ✅ |
| y=three>2 i.eastward., y\epsilon (ii,\infty) | +ve | +ve | +ve i.e., >0 | ❌ |
i.eastward., y=-2, y\epsilon (-ii,2) and y=2
i.e., y\epsilon [-2,two]
Step four: Eliminate the values of y
Equally y=\sqrt{four-x^{2}}, a square root function,
then y can non take any negative value i.e., y\geq 0
Therefore y\epsilon [0,two].
Stride v: Write the range
The range of the function f(x)=\sqrt{4-x^{2}} is [0,2] in interval notation.
We tin likewise write the range of the function f(x)=\sqrt{4-10^{two}} equally R(f)={x\epsilon \mathbb{R}:0\leq y \leq ii}
Instance v: Find the range of a function f(x) =\sqrt{ten^{two}-4}.
Solution:
Pace i: First nosotros equate the function with y
y=\sqrt{x^{2}-iv}
Step 2: Then express 10 as a function of y
y=\sqrt{x^{two}-iv}
or, y^{2}=x^{2}-4
or, x^{ii}=y^{2}+4
Step 3: Discover 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}+four\geq 0 for all y\epsilon \mathbb{R}
Pace 4: Eliminate the values of y
Sincey=\sqrt{10^{ii}-4} is a foursquare root function,
therefore y tin can non take whatsoever negative value i.e.,y\geq 0
Step five: Write the Range
The range of f(ten) =\sqrt{10^{two}-4} is (0,\infty).
Instance 6: Detect the range for the foursquare root office
f(x)=iii-\sqrt{x}
Solution:
#three. Find the range of a function with a square root in the denominator
Example vii: Find the range
f(10)=\frac{1}{\sqrt{x-three}}
Solution:
Pace 1:
y=\frac{one}{\sqrt{x-3}}
Step ii:
y=\frac{1}{\sqrt{x-3}}
or, y=\frac{i}{\sqrt{x-3}}
or, y^{ii}=\frac{1}{x-iii}
or, y^{2}(x-3)=1
or, xy^{ii}-3y^{2}=1
or, xy^{2}=ane+3y^{2}
or, 10=\frac{1+3y^{two}}{y^{ii}}
Step three:
For x=\frac{one+3y^{2}}{y^{2}} to be defined,
y^{2}\neq 0
i.e., y\neq 0
Step iv:
As f(x)=\frac{ane}{\sqrt{x-3}}, so y can non be negative (-ve).
Step v:
The range of f(x)=\frac{i}{\sqrt{x-3}} is (0,\infty).
Instance 8: Discover the range
f(ten)=\frac{1}{\sqrt{4-10^{two}}}
Solution:
Pace ane:
y=\frac{ane}{\sqrt{4-x^{two}}}
Footstep 2:
y=\frac{one}{\sqrt{iv-x^{ii}}}
or, y=\frac{i}{\sqrt{4-x^{2}}}
or, y^{2}=\frac{ane}{iv-10^{2}}
or, 4y^{ii}-x^{ii}y^{2}=1
or, x^{two}y^{two}=4y^{2}-1
or, 10^{2}=\frac{4y^{2}-1}{y^{ii}}
Step 3:
For 10^{ii}=\frac{4y^{2}-1}{y^{2}} to exist divers, y can not be equal to zero
i.e., y\neq 0
Also since x^{two}\geq 0,
\therefore \frac{4y^{ii}-1}{y^{2}}\geq 0
or, 4y^{2}-ane\geq 0 (\because y^{2}\geq 0)
or, (2y-1)(2y+1)\geq 0
or, four(y-\frac{1}{2})(y+\frac{1}{2})\geq 0
| Value of y | Sign of (2y-one) | Sign of (2y+1) | Sign of (2y-ane)(2y+i) | (2y-1)(2y+ane)\geq 0 is satisfied or non |
|---|---|---|---|---|
| y=-i<-\frac{1}{two} i.eastward., y\epsilon \left ( -\infty,-\frac{1}{2} \right ) | -ve | -ve | +ve i.e., >0 | ✅ |
| y=-\frac{1}{2} | -ve | 0 | 0 | ✅ |
| y=0 i.due east., y\epsilon \left (-\frac{i}{2},\frac{ane}{2} \right ) | -ve | +ve | -ve i.e., <0 | ❌ |
| y=\frac{1}{ii} | 0 | +ve | +ve i.eastward., >0 | ✅ |
| y=i>\frac{ane}{2} i.e., y\epsilon \left (\frac{1}{2},\infty \right ) | +ve | +ve | +ve i.east., >0 | ✅ |
The above table implies that
y\epsilon \left ( -\infty,-\frac{1}{ii} \correct )\cup \left (\frac{i}{2},\infty \right ) …..(i)
Pace 4:
Since y=\frac{ane}{\sqrt{iv-ten^{two}}} is a square root office,
therefore y tin can not exist negative (-ve).
i.e., y\geq 0 …..(two)
Now from (1) and (two), nosotros go
y\epsilon ( \frac{ane}{2},\infty )
Step 5:
Therefore the range of the function f(10)=\frac{1}{\sqrt{iv-x^{2}}} is [ \frac{1}{2},\infty )
Example ix: Find the range of the function
f(ten)=\sqrt{\frac{(x-3)(x+2)}{x-1}}
Solution:
#4. Find the range of modulus function or absolute value function
Example ten: Discover the range of the accented value part
f(x)=\left | 10 \right |
Solution:
Nosotros tin can notice the range of the absolute value function f(x)=\left | x \right | on a graph.
If we draw the graph and so we get
Hither 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 | ten \right | is [0,\infty).
Example 11: Detect the range of the absolute value function
f(ten)=-\left | 10-one \correct |
Solution:
The graph of f(ten)=-\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 exist [a,\infty),
- If the sign earlier modulus is negative (-ve) i.eastward., of the form -\left | 10-a \right |, so the range will be (-\infty,a].
We tin besides observe the range of the absolute value functions f(ten)=\left | x \correct | and f(x)=-\left | x-1 \right | using the above short cut play tricks:
The part f(x)=\left | x \right | can exist written as f(10)=+\left | 10-0 \right |
Now using trick ane we can say, the range of f(x)=\left | 10 \right | is [0,\infty)
Also using play a trick on 2 we tin can say, the range of f(x)=-\left | ten-1 \right | is (-\infty,0].
Example 12: Find the range of the following absolute value functions
- f(x)=\left | x \right |+6,
- f(10)=\left | x+four \right |
Solution:
#five. Notice the range of a Stride office
Example 13: Find the range of the step part f(10)=[x],x\epsilon \mathbb{R}.
Solution:
The step function f(x)=[x],x\epsilon \mathbb{R} is expressed as
f(x)=0, 0\leq 10<ane
=1, one\leq 10<ii
=2,2\leq x<3
………
=-1,-ane\leq x<0
=-2,-two\leq x<-1
………
Y'all tin verify this upshot from the graph of f(x)=[x],x\epsilon \mathbb{R}
i.e., y\epsilon {…,-2,-i,0,1,ii,…}
i.due east., y\epsilon \mathbb{Z}, the set up of all integers.
\therefore the range of the step role f(10)=[x],x\epsilon \mathbb{R} is \mathbb{Z}, the set of all integers.
Example xiv: Find the range of the step office f(x)=[ten-3],x\epsilon \mathbb{R}.
Solution:
By using the definition of step function, we can express f(x)=[x-iii],x\epsilon \mathbb{R} as
f(x)=1,3\leq x<iv
=2,iv\leq 10<5
=3,5\leq x<half dozen
………
=0,2\leq 10<3
=-1,1\leq x<2
=-2, 0\leq ten<1
=-three, -1\leq ten<0
………
You can verify this result from the graph of f(x)=[10-3],10\epsilon \mathbb{R}
i.due east., y\epsilon {…,-3,-2,-1,0,one,2,3,…}
i.east., y\epsilon \mathbb{Z}, the set of all integers.
\therefore the range of the footstep function f(x)=[x-3],x\epsilon \mathbb{R} is \mathbb{Z}, the set of all integers.
Example 15: Observe the range of the step office f(x)=\left [ \frac{one}{4x} \right ],x\epsilon \mathbb{R}.
Solution:
#6. Discover the range of an Exponential function
Example 16: Notice the range of the exponential office f(x)=2^{x}.
Solution:
The graph of the role f(10)=two^{ten} is
Here y=0 is an asymptote of f(x)=2^{x} i.e., the graph is going very shut and close to the y=0 straight line but it will never touch y=0.
Also, you lot tin see on the graph that the function is extended to +\infty.
So nosotros can say y>0.
\therefore the range of the exponential role f(x)=ii^{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(ten)=-3^{x+ane}+2 you lot tin can see that y=2 is an asymptote of f(x)=-3^{ten+1}+ii i.due east., on the graph f(10)=-3^{x+ane}+2 is going very close and shut to y=2 towards -ve x-axis but it volition never bear on the direct line y=two and extended to -\infty towards +ve x-axis.
i.e., y<2
\therefore the range of the exponential function f(10)=-3^{ten+1}+two is (-\infty,2).
There is a shortcut play a trick on to find the range of whatever exponential office. This flim-flam will aid y'all find the range of any exponential function in simply ii seconds.
Shortcut trick:
Let f(x)=a\times b^{x-h}+k be an exponential role.
Then
- If a>0, so R(f)=(g,\infty),
- If a<0, then R(f)=(-\infty,m).
Now nosotros effort to find the range of the exponential functions f(x)=2^{10} and f(x)=-3^{x+ane}+ii with the above shortcut trick:
Nosotros tin can write f(x)=ii^{x} as f(ten)=1\times 2^{ten}+0, i>0 and comparing this result with play tricks ane we directly say
The range of f(x)=2^{x} is (0,\infty).
As well f(x)=-three^{x+i}+2 tin be written as f(x)=-one\times three^{x+1}+2, -1<0 and comparing with fob 2 we get
The range of f(10)=-three^{ten+i}+two is (-\infty,two).
Example xviii: Observe the range of the exponential functions given beneath
f(ten)=-two^{x+ane}+3
Solution:
#7. Observe the range of a Logarithmic role
The range of any logarithmic function is (-\infty,\infty).
We can verify this fact from the graph.
f(ten)=\log_{2}x^{3} is a logarithmic function and the graph of this function is
Here you can come across that the y value starts from -\infty and extended to +\infty,
i.east., the range of f(10)=\log_{2}ten^{3} is (-\infty,\infty).
Case 19: Observe the range of the logarithmic function
f(x)=\log_{2}(x+iv)+iii
Solution:
#8. Discover the range of a function relation of ordered pairs
A relation is the set of ordered pairs i.e., the gear up 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 accept learned how to find the domain of a function using relation.
Now we larn how to find the range of a office using relation.
For that nosotros accept to recollect 2 rules which are given below:
Rules:
- Before finding the range of a function first we check the given relation (i.eastward., the set of ordered pairs) is a role or not
- Find all the y values and course a set. This set is the range of the relation.
Now see the examples given below to empathize this concept:
Example 20: Find the range of the relation
{(1,three), (5,9), (8,23), (12,14)}
Solution:
In the relation {(one,3), (5,9), (8,23), (12,14)}, the set of x coordinates is {i, five, 8, 12} and the prepare of y coordinates is {three, 9, 14, 23}.
If we depict the diagram of the given relation it will look like this
Here we can clearly see that each element of the set {1, v, viii, 12} is related to a unique element of the set {iii, 9, xiv, 23}.
Therefore the given relation is a Function.
As well, nosotros know that the range of a part relation is the fix of y coordinates.
Therefore the range of the relation {(1,3), (5,ix), (eight,23), (12,xiv)} is the prepare {iii, nine, 14, 23}.
Instance 21: Observe the range of the set of ordered pairs
{(5,2), (7,half dozen), (9,four), (9,xiii), (12,19)}.
Solution:
The diagram of the given relation is
Hither we can see that element 9 is related to two unlike elements and they are iv and thirteen i.e., 9 is non related to a unique element and this goes against the definition of the function.
Therefore the relation {(five,2), (vii,half dozen), (9,4), (9,13), (12,xix)} is non a Function.
Example 22: Determine the range of the relation described by the table
| x | y |
|---|---|
| -i | 3 |
| 3 | -ii |
| 3 | 2 |
| 4 | 8 |
| half dozen | -1 |
Solution:
#nine. Notice the range of a Discrete part
A Discrete Role is a collection of some points on the Cartesian plane and the range of a discrete role is the set of y-coordinates of the points.
Case 23: How practise you find the range of the discrete function from the graph
Solution:
From the graph, we tin can run into that there are five points on the discrete office and they are A (ii,2), B (4,4), C (six,6), D (eight,8), and E (x,x).
The fix of the y-coordinates of the points A, B, C, D, and Eastward is {ii,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 part is made of the five points A (-3,2), B (-ii,4), C (2,3), D (3,i), and E (5,v).
The ready of the y coordinates of the discrete function is {two,4,3,1,5} = {1,2,three,4,5}.
\therefore the range of the discrete function is {1,ii,3,four,v}.
#x. Discover the range of a trigonometric office
| Trigonometric Office | Expresion | Range |
|---|---|---|
| Sine office | \sin x | [-ane,1] |
| Cosine function | \cos ten | [-one,1] |
| Tangent office | \tan x | (-\infty,+\infty) |
| CSC function (Cosecant function) | \csc x | (-\infty,-1]\cup[1,+\infty) |
| Secant function | \sec 10 | (-\infty,-1]\cup[1,+\infty) |
| Cotangent function | \sec 10 | (-\infty,+\infty) |
#11. Observe the range of an inverse trigonometric function
| Inverse trigonometric function | Expression | Range |
|---|---|---|
| Arc Sine function / Changed Sine function | \arcsin 10 or, \sin^{-ane}x | [-\frac{\pi}{2},+\frac{\pi}{two}] |
| Arc Cosine role / Inverse Cosine function | \arccos ten or, \cos^{-1}x | [0,\pi] |
| Arc Tangent role / 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^{-one}x | [-\frac{\pi}{2},0)\loving 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 part / Inverse Cotangent function | \textrm{arccot}x or, \cot^{-i}x | (0,\pi) |
#12. Detect the range of a hyperbolic part
| Hyperbolic function | Expression | Range |
|---|---|---|
| Hyperbolic Sine function | \sinh ten=\frac{eastward^{x}-e^{-x}}{2} | (-\infty,+\infty) |
| Hyperbolic Cosine role | \cosh 10=\frac{e^{x}+e^{-x}}{two} | [1,\infty) |
| Hyperbolic Tangent function | \tanh x=\frac{due east^{ten}-e^{-x}}{e^{x}+e^{-10}} | (-1,+1) |
| Hyperbolic CSC function | csch x=\frac{2}{e^{ten}-east^{-x}} | (-\infty,0)\loving cup(0,\infty) |
| Hyperbolic Secant function | sech x=\frac{2}{e^{x}+e^{-10}} | (0,1) |
| Hyperbolic Cotangent function | \tanh ten=\frac{e^{x}+e^{-ten}}{e^{10}-e^{-x}} | (-\infty,-1)\loving cup(ane,\infty) |
#13. Notice 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 part | \cosh^{-1}10=\ln(x+\sqrt{x^{2}-1}) | [0,\infty) |
| Changed hyperbolic tangent function | \tanh^{-1}x=\frac{i}{2}\ln\left (\frac{1+x}{one-x}\correct ) | (-\infty,\infty) |
| Changed hyperbolic CSC part | csch^{-1}x=\ln \left ( \frac{ane+\sqrt{1+10^{2}}}{x} \right ) | (-\infty,0)\cup(0,\infty) |
| Inverse hyperbolic Secant function | sech^{-1}x=\ln \left ( \frac{1+\sqrt{ane-ten^{2}}}{10} \correct ) | [0,\infty) |
| Changed hyperbolic Cotangent function | coth^{-1}x=\frac{1}{2}\ln\left (\frac{x+1}{x-ane}\correct ) | (-\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 role consists of two part:
- f(x)=x-3 when x\leq -1,
- f(x)=10+1 when ten>i.
If we plot these two functions on the graph and then we get,
This is the graph of the piecewise office.
From the graph, we can run into that
- the range of the function f(x)=x-three is (-\infty,-two] when x\leq -1,
- the range of the part f(x)=x+1 is (2,\infty) when 10>1,
Therefore from the above results nosotros can say that
The range of the piecewise office f(ten) is
(-\infty,-2]\cup (2,\infty).
Example 26: Observe the range of a piecewise function given beneath
Solution:
If you notice the piecewise office so you can meet there are functions:
- f(ten)=10 divers when 10\leq -one,
- f(10)=2 divers when -1<ten<1),
- f(10)=\sqrt{x} defined when x\geq 1.
At present if we draw the graph of these 3 functions we get,
This is the graph of the piecewise part.
Here y'all can see that
The function f(ten)=ten starts y=-1 and extended to -\infty when x\leq -1.
So the range of the office f(x)=x,x\leq -1 is (-\infty,-ane]……..(1)
The functional value of the function f(x)=2, -ane<x<1 is 2.
The range of the function f(x)=x is {2}……..(2)
The function f(x)=\sqrt{ten} starts at y=1 and extended to \infty when ten\geq one.
The range of the function f(10)=\sqrt{x} is [1,\infty) when 10\geq 1……..(three)
From (ane), (two), and (iii), we get,
the range of the piecewise part is
(-\infty,-ane]\cup {2}\cup [i,\infty)
= (-\infty,-ane]\cup [1,\infty)
#15. Notice the range of a composite part
Instance 27: Let f(x)=2x-6 and m(x)=\sqrt{x} be ii functions.
Find the range of the post-obit composite functions:
(a) f\circ one thousand(x)
(b) one thousand\circ f(10)
Solution of (a)
First we need to find the function g\circ f(x).
We know that,
f\circ yard(x)
=f(k(ten))
=f(\sqrt{x}) (\because g(10)=\sqrt{x})
=two\sqrt{10}-half-dozen
Now meet that ii\sqrt{x}-6 is a part with a foursquare root and at the beginning of this article, nosotros already learned how to notice the range of a function with a foursquare root.
Following these steps, we tin get,
the range of the blended function f of m is
R(f\circ g)=[-6,\infty).
Solution of (b):
grand\circ f(x)
=g(f(ten))
=g(2x-6) (\because f(x)=2x-6)
=\sqrt{2x-half-dozen}, a function with a square root
Using the previous method we get,
the range of the blended function g\circ f(x) is
R(yard\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 blended functions
- f\circ g(ten),
- grand\circ f(x)
Solution:
Anil Kumar (Elapsing: 3 minutes 35 seconds)
Also read:
- How to Notice the Domain of a Part Algebraically – All-time ix Means
- 3 ways to find the zeros of a function
- How to observe the zeros of a quadratic function?
- thirteen ways to find the limit of a role
- How to use the Squeeze theorem to find a limit?
Source: https://mathculus.com/how-to-find-the-range-of-a-function-algebraically/
Posted by: millsextre1971.blogspot.com
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