chapter 7 radicals, radical functions, and rational exponents
DESCRIPTION
Chapter 7 Radicals, Radical Functions, and Rational Exponents. § 7.1. Radical Expressions and Functions. 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 - PowerPoint PPT PresentationTRANSCRIPT
Chapter 7Radicals, Radical
Functions, and Rational Exponents
§ 7.1
Radical Expressions and Functions
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 7.1
RadicalsIn this section, we introduce a new category of expressions and functionsthat contain roots.
For example, the reverse operation of squaring a number is finding thesquare 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 aradical expression.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 7.1
Radical ExpressionsEXAMPLE
n a
Radical Expression
Radicand
Radical Sign
Index of the Radical
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 7.1
Radical Expressions
Definition of the Principal Square RootIf a is a nonnegative real number, the nonnegative number b such that , denoted by , is the principal square root of a.
ab 2 ab
P 487
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 7.1
Radical ExpressionsEXAMPLE
Evaluate: .25144(c)25144(b)16(a)
SOLUTIONThe principal square root of a negative number, -16, is not a real number.Simplify the radicand. The principal square root of 169 is 13.
16(a)
1316925144(b)
1751225144(c) Take the principal square root of 144, 12, and of 25, 5, and then add to get 17.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 7.1
Radical ExpressionsCheck Point 1 on page 487
Evaluate:64
49-
525169 743169
87
2516
54
0081.0 09.0
Principal Square Root means the answer is nonnegative
- Denotes the negative square root of a number
Is a grouping symbol
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 7.1
Radical Expressions
Square Root FunctionBecause each nonnegative real number, x, has precisely one principal square root, , there is a square root function defined by The domain of this function is .
xxf )(
x
Bottom of P 487
),0[
x (x,y)0 (0,0)
1 (1,1)
4 (4,2)
9 (9,3)
16 (16,4)
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 7.1
Radical Expressions
xxf )(
P 488
00)0( f
11)1( f
24)4( f
39)9( f
416)16( f
See Figure 7.1 on page 488
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 7.1
Radical FunctionsEXAMPLE
For the function, find the indicated function value:
.1,21,1,4;12
ggggxxg
SOLUTION
Substitute 4 for x in 39
1424
g
73.13
1121
g
.12 xxg
Simplify the radicand and take the square root of 9.
Substitute 1 for x in .12 xxg
Simplify the radicand and take the square root of 3.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 7.1
Radical Functions
Substitute -1/2 for x in
00
1212
21
g
1
1121g
.12 xxgSimplify the radicand and take the square root.
Substitute -1 for x in .12 xxgSimplify the radicand. The principal square root of a negative number is not a real number.
CONTINUED
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 7.1
Radical FunctionsCheck Point 2
For the function, find the indicated function value:
xxgxxf 39;2012
SOLUTION
3f
5g
4
90.424
P 488
20312
)5(39
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 7.1
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.
In other words, we only use “allowable” x in the domain of the function. Not allowed for x is any value of x that would cause a negative number under a square root.
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 7.1
Radical Functions - DomainEXAMPLE
Find the domain of .153 xxf
SOLUTIONThe 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.
The domain of f is . ,5or 5| xx
0153 x153 x5x
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 7.1
Radical FunctionsCheck Point 3
Find the Domain of
279 xxf
SOLUTION
),3[or 3| : xxxfdomain
P 489
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 7.1
Radical Functions in ApplicationEXAMPLE
Police use the function 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.
xxf 20
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 7.1
Radical Functions in Application
SOLUTION
Use the given function. xxf 20
CONTINUED
Substitute 45 for x. 4520xf
Simplify the radicand. 900xf
Take the square root. 30xf
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.
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 7.1
Radical Expressions
Simplifying TFor any real number a,
In words, the principal square root of is the absolute value of a.
.2 aa
2a
2a
The principal root is the positive root. P 490
(a) To simplify , first write as an expression that is squared: . Then simplify.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 7.1
Radical ExpressionsEXAMPLE
Simplify each expression:
.4914(b)81(a) 24 xxx
SOLUTIONThe 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.
481x 481x 224 981 xx
22224 9or 9981 xxxx
(b) To simplify , first write as an expression that is squared: . Then simplify.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 7.1
Radical Expressions
49142 xx
CONTINUED
49142 xx 22 74914 xxx
774914 22 xxxx
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 7.1
Radical FunctionsCheck Point 5
Simplify each expression:
7
8 x
P 490
2)7( . a
2)8( . xb
1049 . xc57x
96 . 2 xxd 3 x
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 7.1
Radical Expressions
Definition of the Cube Root of a NumberThe cube root of a real number a is written .
. that means 33 abba
3 a
P 491
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 7.1
Radical FunctionsEXAMPLE
For the function, find the indicated function value:
.63,0,13;123 gggxxg
SOLUTION
Substitute 13 for x in 327
1132133
3
g
11
1020 3
g
.123 xxg
Simplify the radicand and take the cube root of 27.
Substitute 0 for x in .123 xxg
Simplify the radicand and take the cube root of 1.
Blitzer, Intermediate Algebra, 5e – Slide #24 Section 7.1
Radical Functions
Substitute -63 for x in 55125
1632633
3
g .123 xxg
Simplify the radicand and take the cube root of -125 and then simplify.
CONTINUED
x (x,y)-8 (-8,-2)
-1 (-1,-1)
0 (0,0)
1 (1,1)
8 (8,2)
Blitzer, Intermediate Algebra, 5e – Slide #25 Section 7.1
Radical Expressions
3)( xxf
P 491
28)8( 3 f
See Figure 7.2 on page 491
11)1( 3 f
00)0( 3 f
11)1( 3 f
28)8( 3 f
Blitzer, Intermediate Algebra, 5e – Slide #26 Section 7.1
Radical FunctionsCheck Point 6
For the function, find the indicated function value:
3 6 . xxfa
33f
5g
3
2
P 492
3 633
3 2)5(2
3 22 . xxgb
3 27
3 8
Blitzer, Intermediate Algebra, 5e – Slide #27 Section 7.1
Radical Expressions
Simplifying TFor any real number a,
In words, the cube root of any expression is that expression cubed.
.3 3 aa
3 3a
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 7.1
Radical ExpressionsEXAMPLE
Simplify: .1253 3x
SOLUTION
Begin by expressing the radicand as an expression that is cubed: . Then simplify.
We can check our answer by cubing -5x:
33 5125 xx
xxx 55125 3 33 3
3333 12555 xxx
By obtaining the original radicand, we know that our simplification is correct.
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 7.1
Radical FunctionsCheck Point 7
Simplify:
x3
P 492
3 327x
Blitzer, Intermediate Algebra, 5e – Slide #30 Section 7.1
Radical ExpressionsEXAMPLE
Find the indicated root, or state that the expression is not a real number:
.1(b)1 (a) 85
SOLUTION
is not a real number because the index, 8, is even and the radicand, -1, is negative. No real number can be raised to the eighth power to give a negative result such as -1. Real numbers to even powers can only result in nonnegative numbers.
because . An odd root of a negative real number is always negative.
11 (a) 5 1111111 5
8 1(b)
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 7.1
Radical Expressions
Simplifying TFor any real number a,
1) If n is even,
2) If n is odd,. aan n
n na
. aan n
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 7.1
Radical ExpressionsEXAMPLE
Simplify: .232(b)5(a) 5 54 4 xx
SOLUTIONEach 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.
if n is even. 55(a) 4 4 xx aan n
2222
22232(b)
5 5
5 555 5
xx
xx if n is odd.aan n
Blitzer, Intermediate Algebra, 5e – Slide #33 Section 7.1
Radical FunctionsCheck Point 8
Find the indicated root, or state not real:
2
2
P 493
4 16 .a
4 16- .b
4 16 . c realnot
5 1- .d 1
Blitzer, Intermediate Algebra, 5e – Slide #34 Section 7.1
Radical FunctionsCheck Point 9
Simply:
6 x
P 494
4 46 . xa
6 6)8( . c 8
5 52-3x .b 23 x
DONE