Download 7.1 - CSE

Section 7.1
Radical
Expressions
and Functions
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1
Objective #1
Evaluate square roots.
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Radicals
In this section, we introduce a new category of expressions and functions
that contain roots.
For example, the reverse operation of squaring a number is finding the
square root of the number.
The symbol
that we use to denote the principal square root
is called a radical sign. The number under the radical sign is called the
radicand. Together we refer to the radical sign and its radicand as a
radical expression.
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Radical Expressions
EXAMPLE
Index
of the
Radical
n
a
Radicand
Radical
Sign
Radical
Expression
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Radical Expressions
Definition of the Principal Square Root
If a is a nonnegative real number, the nonnegative
number b such that b 2 = a , denoted by b = a , is the
principal square root of a.
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Radical Expressions
EXAMPLE
Evaluate: (a) − 16 (b) 144 + 25 (c) 144 + 25.
SOLUTION
(a) −16 is not a real number.
(b) 144 + 25 = 169 = 13
Simplify the radicand. The
principal square root of 169 is
13.
(c) 144 + 25 = 12 + 5 = 17
Take the principal square root
of 144, 12, and of 25, 5, and
then add to get 17.
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Objective #1: Example
1a. Evaluate:
64
64 = 8 because 82 = 64 .
1b. Evaluate: − 49
− 49 =
−7 because ( −7) 2 =
49 .
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Objective #1: Example
1a. Evaluate:
64
64 = 8 because 82 = 64 .
1b. Evaluate: − 49
− 49 =
−7 because ( −7) 2 =
49 .
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Objective #1: Example
1c. Evaluate:
16
25
2
16 4
4
16
 
= because
= .
 5
25 5
25
1d. Evaluate:
9 + 16
9 + 16 =25
=5
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Objective #1: Example
1c. Evaluate:
16
25
2
16 4
4
16
 
= because
= .
 5
25 5
25
1d. Evaluate:
9 + 16
9 + 16 =25
=5
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Objective #2
Evaluate square root functions.
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Radical Functions
EXAMPLE
For the function, find the indicated function value:
 1
g (x ) = − 2 x + 1; g (4 ), g (1), g  − , g (− 1).
 2
SOLUTION
g (4 ) = − 2(4 ) + 1
= − 9 = −3
g (1) = − 2(1) + 1
= − 3 = −1.73
Substitute 4 for x in g (x ) = − 2 x + 1.
Simplify the radicand and take the
square root of 9.
Substitute 1 for x in g (x ) = − 2 x + 1.
Simplify the radicand and take the
square root of 3.
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Radical Functions
CONTINUED
 1
 1
g  −  = − 2 −  + 1
 2
 2
=− 0 =0
g (− 1) = − 2(− 1) + 1
= − −1 = ∅
g ( x ) = − 2 x + 1.
Simplify the radicand and take the
square root.
g ( x ) = − 2 x + 1.
Simplify the radicand. The principal
square root of a negative number is
not a real number.
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Objective #2: Example
2a. If =
f ( x)
=
f ( x)
12 x − 20, find f (3).
=
f (3)
12 x − 20
=
12(3) − 20
36 − 20
= 16
=4
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Objective #2: Example
2a. If =
f ( x)
=
f ( x)
12 x − 20, find f (3).
=
f (3)
12 x − 20
=
12(3) − 20
36 − 20
= 16
=4
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Objective #2: Example
2b. If g ( x) =
− 9 − 3 x , find g ( −5).
g ( x) =
− 9 − 3x
g ( −5) =− 9 − 3( −5)
=
− 9 + 15
= − 24
≈ −4.90
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Objective #2: Example
2b. If g ( x) =
− 9 − 3 x , find g ( −5).
g ( x) =
− 9 − 3x
g ( −5) =− 9 − 3( −5)
=
− 9 + 15
= − 24
≈ −4.90
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Objective #3
Find the domain of square root functions.
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Radical Functions - Domain
We have seen that the domain of a function f is the largest set of real
numbers for which the value of f(x) is defined.
Because only nonnegative numbers have real square roots, the
domain of a square root function is the set of real numbers for which
the radicand is nonnegative.
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Radical Functions - Domain
EXAMPLE
Find the domain of f (x ) = 3x − 15.
SOLUTION
The domain is the set of real numbers, x, for which the radicand,
3x – 15, is nonnegative. We set the radicand greater than or
equal to 0 and solve the resulting inequality.
3 x − 15 ≥ 0
3 x ≥ 15
x≥5
The domain of f is {x | x ≥ 5} or [5, ∞ ) .
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Objective #3: Example
3. Find the domain of f =
( x)
9 x − 27 .
The radicand, 9 x − 27, must be nonnegative.
9 x − 27 ≥ 0
9 x ≥ 27
x≥3
The domain of f is [3, ∞) .
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Objective #3: Example
3. Find the domain of f =
( x)
9 x − 27 .
The radicand, 9 x − 27, must be nonnegative.
9 x − 27 ≥ 0
9 x ≥ 27
x≥3
The domain of f is [3, ∞) .
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Objective #4
Use models that are square root functions.
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Radical Functions in Application
The graph of the square root function f ( x ) = x is increasing
from left to right. However, the rate of increase is slowing down
as the graph moves to the right. This is why square root
functions are often used to model growing population with
growth that is leveling off.
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Radical Functions in Application
EXAMPLE
Police use the function f (x ) = 20 x to estimate the speed of a
car, f (x), in miles per hour, based on the length, x, in feet, of its
skid marks upon sudden braking on a dry asphalt road. Use the
function to solve the following problem.
A motorist is involved in an accident. A police officer measures
the car’s skid marks to be 45 feet long. Estimate the speed at
which the motorist was traveling before braking. If the posted
speed limit is 35 miles per hour and the motorist tells the officer
she was not speeding, should the officer believe her? Explain.
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Radical Functions in Application
CONTINUED
SOLUTION
f ( x ) = 20 x
Use the given function.
f ( x ) = 20(45)
Substitute 45 for x.
f ( x ) = 900
Simplify the radicand.
f ( x ) = 30
Take the square root.
The model indicates that the motorist was traveling at 30 miles
per hour at the time of the sudden braking. Since the posted
speed limit was 35 miles per hour, the officer should believe that
she was not speeding.
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Objective #4: Example
4. An “hour-long” TV show is actually much shorter when you
consider the amount of “clutter,” including commercials and
plugs for other shows. The square root function
M
=
( x) 0.7 x + 12.5 models the average number of
nonprogram minutes, M ( x), in an hour of prime-time cable x
years after 1996. Use the function to predict how many cluttered
minutes, rounded to the nearest tenth, there will be in an hour in
2014.
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Objective #4: Example
CONTINUED
Because 2014 is 18 years after 1996, substitute 18 for x.
M
=
( x) 0.7 x + 12.5
M
=
(18) 0.7 18 + 12.5
≈ 15.5
The model indicates that there will be about 15.5 nonprogram
minutes in an hour of prime-time cable in 2014.
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Objective #4: Example
CONTINUED
Because 2014 is 18 years after 1996, substitute 18 for x.
M
=
( x) 0.7 x + 12.5
M
=
(18) 0.7 18 + 12.5
≈ 15.5
The model indicates that there will be about 15.5 nonprogram
minutes in an hour of prime-time cable in 2014.
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Objective #5
Simplify expressions of the form a 2 .
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Radical Expressions
Simplifying a 2 T
For any real number a,
a2 = a .
2
In words, the principal square root of a is the absolute
value of a.
The principal root is the positive root.
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Radical Expressions
EXAMPLE
Simplify each expression: (a)
81x 4
(b)
x 2 + 14 x + 49 .
SOLUTION
The principal square root of an expression squared is the
absolute value of that expression. In both exercises, it will first
be necessary to express the radicand as an expression that is
squared.
(a) To simplify 81x 4, first write 81x 4 as an expression that is
2
squared: 81x 4 = (9 x 2 ) . Then simplify.
81x =
4
(9 x )
2 2
= 9 x 2 or 9 x 2
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Radical Expressions
CONTINUED
(b) To simplify x 2 + 14 x + 49, first write x 2 + 14 x + 49 as an
2
expression that is squared: x 2 + 14 x + 49 = (x + 7 ) . Then simplify.
x 2 + 14 x + 49 =
(x + 7 )2
= x+7
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Objective #5: Example
5a. Simplify:
( −7) 2
( −7) 2 =
49
=7
5b. Simplify:
( x + 8) 2
( x + 8) 2 =x + 8
5c. Simplify:
49x10
49 x10 = 7 x5
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Objective #5: Example
5a. Simplify:
( −7) 2
( −7) 2 =
49
=7
5b. Simplify:
( x + 8) 2
( x + 8) 2 =x + 8
5c. Simplify:
49x10
49 x10 = 7 x5
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Objective #6
Evaluate cube root functions.
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Radical Expressions
Definition of the Cube Root of a Number
The cube root of a real number a is written
3
3
.a
a = b means that b 3 = a.
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Radical Functions
EXAMPLE
For the function, find the indicated function value:
g ( x ) = −3 2 x + 1; g (13), g (0 ), g (− 63).
SOLUTION
g (13) = −3 2(13) + 1
= −3 27 = −3
g (0 ) = − 2(0 ) + 1
3
= − 1 = −1
Substitute 13 for x in g (x ) = −3 2 x + 1.
Simplify the radicand and take the
cube root of 27.
Substitute 0 for x in g (x ) = −3 2 x + 1.
Simplify the radicand and take the
cube root of 1.
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Radical Functions
CONTINUED
g (− 63) = −3 2(− 63) + 1
g ( x ) = −3 2 x + 1.
= −3 − 125 = −(− 5) = 5
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Objective #6: Example
6a. If f (=
x) 3 x − 6, find f (33).
f (=
x)
3
x−6
f (33)
=
3
33 − 6
= 3 27
=3
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Objective #6: Example
6a. If f (=
x) 3 x − 6, find f (33).
f (=
x)
3
x−6
f (33)
=
3
33 − 6
= 3 27
=3
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Objective #6: Example
6b. If g=
( x) 3 2 x + 2, find g ( −5).
g=
( x)
3
2x + 2
g ( −5)=
3
2( −5) + 2
=
3
−8
= −2
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Objective #6: Example
6b. If g=
( x) 3 2 x + 2, find g ( −5).
g=
( x)
3
2x + 2
g ( −5)=
3
2( −5) + 2
=
3
−8
= −2
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Objective #7
3 3
Simplify expressions of the form a .
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Radical Expressions
Simplifying
3
a3 T
For any real number a,
3
a 3 = a.
In words, the cube root of any expression cubed is that
expression.
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Radical Expressions
EXAMPLE
Simplify: 3 − 125 x 3 .
SOLUTION
Begin by expressing the radicand as an expression that is cubed:
3
− 125 x 3 = (− 5 x ) . Then simplify.
3
− 125 x 3 = 3 (− 5 x ) = −5 x
3
(− 5 x )3 = (− 5)3 x 3 = −125 x 3
By obtaining the original radicand, we know that our
simplification is correct.
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Objective #7: Example
7. Simplify:
3
3
−27x3
3
( −3 x)3
−27 x3 =
= −3 x
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Objective #7: Example
7. Simplify:
3
3
−27x3
3
( −3 x)3
−27 x3 =
= −3 x
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Objective #8
Find even and odd roots.
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Radical Expressions
EXAMPLE
Find the indicated root, or state that the expression is not a real
number:
(a) 5 − 1 (b) 8 − 1.
SOLUTION
(a) 5 − 1 = −1 because (− 1)5 = (− 1)(− 1)(− 1)(− 1)(− 1) = −1 . An odd root
of a negative real number is always negative.
(b) 8 − 1
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Objective #8: Example
8a. Find the indicated root, or state that the expression is not a real
number: − 4 16.
− 4 16 =
−2
8b. Find the indicated root, or state that the expression is not a real
number: 4 −16.
4
−16 is not a real number.
8c. Find the indicated root, or state that the expression is not a real
number: 5 −1.
5
−1 =−1
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Objective #8: Example
8a. Find the indicated root, or state that the expression is not a real
number: − 4 16.
− 4 16 =
−2
8b. Find the indicated root, or state that the expression is not a real
number: 4 −16.
4
−16 is not a real number.
8c. Find the indicated root, or state that the expression is not a real
number: 5 −1.
5
−1 =−1
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Objective #9
n
Simplify expressions of the form a n .
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Radical Expressions
Simplifying
n
an
T
For any real number a,
1) If n is even,
n
an = a .
2) If n is odd,
n
an = a .
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Radical Expressions
EXAMPLE
4
Simplify: (a) 4 (x + 5)
(b) 5 − 32( x − 2 ) .
5
SOLUTION
Each expression involves the nth root of a radicand raised to the
nth power. Thus, each radical expression can be simplified.
Absolute value bars are necessary in part (a) because the index,
n, is even.
(a) 4 ( x + 5) = x + 5
4
(b) 5 − 32( x − 2 ) = 5 (− 2 ) ( x − 2 )
5
5
5
n
a n = a if n is even.
n
a n = a if n is odd.
= 5 [− 2( x − 2 )]
5
= −2( x − 2 )
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Objective #9: Example
9a. Simplify:
4
( x + 6) 4
( x + 6) 4 =x + 6
9b. Simplify:
5
4
5
(3 x − 2)5
(3 x − 2)5 =3 x − 2
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Objective #9: Example
9a. Simplify:
4
( x + 6) 4
( x + 6) 4 =x + 6
9b. Simplify:
5
4
5
(3 x − 2)5
(3 x − 2)5 =3 x − 2
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