Transcript College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson P Prerequisites.

College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
P Prerequisites
P.8
Rational Expressions
Fractional Expression
A quotient of two algebraic expressions
is called a fractional expression.
• Here are some examples:
2x
x 1
x 3
x 1
y 2
2
y 4
Rational Expression
A rational expression is a fractional
expression where both the numerator
and denominator are polynomials.
• Here are some examples:
2x
x 1
x
2
x 1
x3  x
2
x  5x  6
Rational Expressions
In this section, we learn:
• How to perform algebraic operations
on rational expressions.
The Domain of
an Algebraic Expression
The Domain of an Algebraic Expression
In general, an algebraic expression may
not be defined for all values of the variable.
The domain of an algebraic expression is:
• The set of real numbers that the variable
is permitted to have.
The Domain of an Algebraic Expression
The table gives some basic expressions
and their domains.
E.g. 1—Finding the Domain of an Expression
Consider the expression
2x  4
x 3
a) Find the value of the expression for
x = 2.
b) Find the domain of the expression.
E.g. 1—The Value of an Expression Example (a)
We find the value by substituting 2 for x in the
expression:
2(2)  4
 8
23
E.g. 1—Domain of an Expression
Example (b)
The denominator is zero when x = 3.
Since division by zero is not defined:
• We have x ≠ 3.
• Thus, the domain is all real numbers except 3.
• We can write this in set notation as
{x | x ≠ 3}
E.g. 2—Finding the Domain of an Expression
Find the domains of these expressions.
2
a
2
x
 3x  1
 
x
b  2
x  5x  6
x
c 
x 5
E.g. 2—Finding the Domain
Example (a)
2x2 + 3x – 1
This polynomial is defined for every x.
• Thus, the domain is the set
of real numbers.
E.g. 2—Finding the Domain
Example (b)
We first factor the denominator.
x
x

2
x  5 x  6  x  2 x  3 
• The denominator is zero when x = 2 or x = 3.
• So, the expression is not defined for these
numbers.
• Hence, the domain is: {x | x ≠ 2 and x ≠ 3}
E.g. 2—Finding the Domain
Example (c)
x
x 5
• For the numerator to be defined,
we must have x ≥ 0.
• Also, we cannot divide by zero; so, x ≠ 5.
• Thus, the domain is:
{x | x ≥ 0 and x ≠ 5}
Simplifying
Rational Expressions
Simplifying Rational Expressions
To simplify rational expressions, we factor
both numerator and denominator and use
this property of fractions:
AC A

BC B
• This allows us to cancel common factors
from the numerator and denominator.
E.g. 3—Simplifying by Cancellation
x

1
Simplify: 2
x  x 2
2
x 1
x2  x  2
x  1 x  1


 x  1 x  2
2
x 1

x2
(Factor)
(Cancel common factors)
Caution
We can’t cancel the x2’s in
x 1
2
x  x 2
2
because x2 is not a factor.
Multiplying and Dividing
Rational Expressions
Multiplying Rational Expressions
To multiply rational expressions, we use
this property of fractions:
A C AC
 
B D BD
This says that:
• To multiply two fractions, we multiply their
numerators and multiply their denominators.
E.g. 4—Multiplying Rational Expressions
Perform the indicated multiplication
and simplify:
x  2 x  3 3 x  12

2
x  8 x  16 x  1
2
E.g. 4—Multiplying Rational Expressions
We first factor.
x 2  2 x  3 3 x  12

2
x  8 x  16 x  1
x  1 x  3  3  x  4 



Factor
2
x 1
 x  4

3  x  1 x  3  x  4 
 x  1 x  4 
3  x  3

x4
2
Property of fractions
Cancel common factors
Dividing Rational Expressions
To divide rational expressions, we use
this property of fractions:
A C A D
  
B D B C
This says that:
• To divide a fraction by another fraction,
we invert the divisor and multiply.
E.g. 5—Dividing Rational Expressions
Perform the indicated division and
simplify:
x  4 x  3x  4
 2
2
x  4 x  5x  6
2
E.g. 5—Dividing Rational Expressions
x  4 x 2  3x  4
 2
2
x  4 x  5x  6
x  4 x 2  5x  6
 2
 2
Invert and multiply
x  4 x  3x  4
x  4  x  2  x  3 


Factor
 x  2  x  2  x  4  x  1
x 3

 x  2  x  1
Cancel common factors
Adding and Subtracting
Rational Expressions
Adding and Subtracting Rational Expressions
To add or subtract rational expressions,
we first find a common denominator and
then use this property of fractions:
A B AB
 
C C
C
Adding and Subtracting Rational Expressions
Any common denominator will work.
Still, it is best to use the least common
denominator (LCD) as learnt in Section P.2.
• The LCD is found by factoring each denominator
and taking the product of the distinct factors,
using the highest power that appears in any
of the factors.
Caution
Avoid making the following error:
A
A A
 
B C B C
Caution
For instance, if we let A = 2, B = 1, and C = 1,
then we see the error:
2 ?2 2
 
1 1 1 1
2?
2  2
2
?
1 4
Wrong!
E.g. 6—Adding and Subtracting Rational Expressions
Perform the indicated operations and
simplify:
3
x
(a)

x 1 x  2
1
2
(b) 2

2
x  1  x  1
E.g. 6—Adding
Example (a)
The LCD is simply the product (x – 1)(x + 2).
3
x

x 1 x  2
3  x  2
x  x  1


 x  1 x  2  x  1 x  2
Fractions by LCD
3x  6  x  x

 x  1 x  2
Add fractions
x  2x  6

 x  1 x  2
Combine terms in numerator
2
2
E.g. 6—Subtracting
Example (b)
The LCD of x2 – 1 = (x – 1)(x + 1) and (x + 1)2
is (x – 1)(x + 1)2.
1
2

2
x  1  x  12

1

2
 x  1 x  1  x  1
x  1  2  x  1


2
 x  1 x  1
2
Factor
Combine using LCD
E.g. 6—Subtracting Rational Exp.


x  1  2x  2
 x  1 x  1
2
3x
 x  1 x  1
2
Example (b)
Distributive Property
Combine terms in numerator
Compound Fractions
Compound Fraction
A compound fraction is:
• A fraction in which the numerator,
the denominator, or both, are themselves
fractional expressions.
E.g. 7—Simplifying a Compound Fraction
Simplify:
x
1
y
y
1
x
E.g. 7—Simplifying
Solution 1
One solution is as follows.
1. Combine the terms in the numerator into
a single fraction.
2. Do the same in the denominator.
3. Invert and multiply.
E.g. 7—Simplifying
Thus,
Solution 1
x
xy
1
y
y

y
xy
1
x
x
xy
x


y
xy

x x  y 
y x  y 
E.g. 7—Simplifying
Solution 2
Another solution is as follows.
1. Find the LCD of all the fractions in
the expression.
2. Multiply the numerator and denominator by it.
E.g. 7—Simplifying
Solution 2
Here, the LCD of all the fractions is xy.
x
x
1
1
xy
y
y


y
y xy
1
1
x
x
2
x  xy

2
xy  y

x x  y 
y x  y 
Multiply num. and denom.by xy
Simplify
Factor
Simplifying a Compound Fraction
The next two examples show
situations in calculus that require
the ability to work with fractional
expressions.
E.g. 8—Simplifying a Compound Fraction
1
1
Simplify:

ah a
h
• We begin by combining the fractions in
the numerator using a common denominator.
E.g. 8—Simplifying a Compound Fraction
1
1

ah a
h
a  a  h 
a a  h 

h
a  a  h  1


a a  h  h
Invert divisor and multiply
E.g. 8—Simplifying a Compound Fraction
aah 1


a a  h  h
DistributiveProperty
h
1


a a  h  h
Simplify
1

a a  h 
Cancel common factors
E.g. 9—Simplifying a Compound Fraction
Simplify:
1  x 
2
1
2
2

1 x
2
 x 1 x

1
2
2
E.g. 9—Simplifying
Solution 1
Factor (1 + x2)–1/2 from the numerator.
1  x 
2
1
2

 x 1 x
2
1 x 2
2

1
2

1
2


1 x 


1
2
1 x
2
2

1 x 2
1

1 x
2



 1 x 2  x 2 


1 x 2
3
2
E.g. 9—Simplifying
Solution 2
(1 + x2)–1/2 = 1/(1 + x2)1/2 is a fraction.
Therefore, we can clear all fractions by
multiplying numerator and denominator
by (1 + x2)1/2.
E.g. 9—Simplifying
Solution 2
Thus,

1 x 2


1
2
 x 2 1 x
1 x
1 x 


2



1 x
2
 x 1 x

1 x 2  x 2
1  x 
2
2
2
1
2

1
2
2
3
2

2

1
2
1
1  x 
2
3
2
 1  x 
1  x 
2
1
2
2
1
2
Rationalizing the Denominator
or the Numerator
Rationalizing the Denominator
If a fraction has a denominator of the form
AB C
we may rationalize the denominator by
multiplying numerator and denominator
by the conjugate radical
AB C
Rationalizing the Denominator
This is effective because, by Special Product
Formula 1 in Section P.6, the product of
the denominator and its conjugate radical
does not contain a radical:

AB C


A  B C  A2  B 2C
E.g. 10—Rationalizing the Denominator
Rationalize the denominator:
1
1 2
• We multiply both the numerator and
the denominator by the conjugate radical
of 1  2 , which is 1  2 .
E.g. 10—Rationalizing the Denominator
Thus,
1
1 2


1

1 2
1 2 1 2
1 2
1 
2
 2
2
Special Product Formula 1
1 2 1 2


 2 1
1 2
1
E.g. 11—Rationalizing the Numerator
Rationalize the numerator:
4h 2
h
• We multiply numerator and denominator
by the conjugate radical 4  h  2 .
E.g. 11—Rationalizing the Numerator
Thus,
4h 2
h
4h 2 4h 2

h
4h 2



h
4h


2
2
2
4h 2

Special Product Formula1
E.g. 10—Rationalizing the Numerator

h

h




4h4
4h 2
h
4h 2
1
4h 2



Cancel common factors
Avoiding Common Errors
Avoiding Common Errors
Don’t make the mistake of
applying properties of multiplication
to the operation of addition.
• Many of the common errors in algebra
involve doing just that.
Avoiding Common Errors
The table states several multiplication
properties and illustrates the error in applying
them to addition.
Avoiding Common Errors
To verify that the equations in the right-hand
column are wrong, simply substitute numbers
for a and b and calculate each side.
Avoiding Common Errors
For example, if we take a = 2 and b = 2
in the fourth error, we have the following
result.
Avoiding Common Errors
1 1 1 1
The left-hand side is:
   1
a b 2 2
1
1
1
The right-hand side is:


ab 22 4
• Since 1 ≠ ¼, the stated equation is wrong.
Avoiding Common Errors
You should similarly convince yourself
of the error in each of the other equations.
• See Exercise 113.