please complete the prerequisite skills pg 412 #1-12

PLEASE COMPLETE THE PREREQUISITE SKILLS

PG 412 #1-12

B IG IDEAS : USE RAT IONAL EXPONENTS

PERFORMING FUNCT ION OPERAT IONS AND F IND ING

INVERSE FUNCT IONS• SOLV ING RAD ICAL EQUAT IONS

CHAPTER 6:RATIONAL EXPONENTS AND

RADICAL FUNCTIONS

LESSON 1: EVALUATE NTH ROOTS AND USE RATIONAL

EXPONENTS

ESSENTIAL QUESTION

WHAT IS THE RELATIONSHIP BETWEEN

NTH ROOTS AND RATIONAL EXPONENTS?

VOCABULARY

• Nth root of a: For an integer n greater than 1, if bn = a, then b is an nth root of a. written as

• Index of a radical: The integer n, greater than 1, in the expression

EXAMPLE 1 Find nth roots

Find the indicated real nth root(s) of a.

a. n = 3, a = –216 b. n = 4, a = 81

SOLUTION

b. Because n = 4 is even and a = 81 > 0, 81 has two real fourth roots. Because 34 = 81 and (–3)4 = 81, you can write ±4√ 81 = ±3

a. Because n = 3 is odd and a = –216 < 0, –216 has one real cube root. Because (–6)3 = –216, you can write = 3√ –216 = –6 or (–216)1/3 = –6.

EXAMPLE 2 Evaluate expressions with rational exponents

Evaluate (a) 163/2 and (b) 32–3/5.

SOLUTION

Rational Exponent Form Radical Forma. 163/2 (161/2)3 = 43= 64= 163/2 ( )3

= 16 43= 64=

b. 32–3/5 = 1323/5

= 1(321/5)3

= 123

18=

32–3/51

323/5= 1( )35 32

=

= 123

18=

EXAMPLE 3 Approximate roots with a calculator

Expression Keystrokes Display

a. 91/5 9 1 5 1.551845574

b. 123/8 12 3 8 2.539176951

7c. ( 4 )3 = 73/4 7 3 4 4.303517071

GUIDED PRACTICE for Examples 1, 2 and 3

Find the indicated real nth root(s) of a.

1. n = 4, a = 625

SOLUTION ±5

2. n = 6, a = 64

SOLUTION ±2

3. n = 3, a = –64.

–4

4. n = 5, a = 243

3SOLUTION

SOLUTION


Evaluate expressions without using a calculator.

5. 45/2

SOLUTION 32

6. 9–1/2

SOLUTION 13

7. 813/4

SOLUTION 27

8. 17/8

SOLUTION 1


Evaluate the expression using a calculator. Round the result to two decimal places when appropriate.

Expression

9. 42/5 1.74

0.06

32

9.65

10. 64 2/3–

11. (4√ 16)5

12. (3√ –30)2

SOLUTION

SOLUTION

SOLUTION

SOLUTION

EXAMPLE 4 Solve equations using nth roots

Solve the equation.

a. 4x5 = 128

Divide each side by 4.x5 32=

Take fifth root of each side.x = 325

Simplify.x 2=

EXAMPLE 4 Solve equations using nth roots

b. (x – 3)4 = 21

Take fourth roots of each side.

x – 3 4+– 21=

Add 3 to each side.x 4+– 21 + 3=

Write solutions separately.

x 421 + 3= or x = 421 + 3–

Use a calculator.x 5.14 or x 0.86

ESSENTIAL QUESTION

WHAT IS THE RELATIONSHIP BETWEEN

NTH ROOTS AND RATIONAL EXPONENTS?

The nth root of a can be written as a to the

SIMPLIFY THE EXPRESSION:43*48

LESSON 2: APPLY PROPERTIES OF RATIONAL EXPONENTS

ESSENTIAL QUESTION

HOW ARE THE PROPERTIES OF

RATIONAL EXPONENTS RELATED TO PROPERTIES

OF INTEGER EXPONENTS?

VOCABULARY

• Simplest form of a radical: A radical with index n is in simplest form if the radicand has no perfect nth powers as factors and any denominator has been rationalized

• Like radicals: Radical expressions with the same index and radicand

EXAMPLE 1 Use properties of exponents

= (71/3)2

= 12 –1

Use the properties of rational exponents to simplify the expression.

b. (61/2 41/3)2 = (61/2)2 (41/3)2 = 61 42/3 = 6 42/3= 6(1/2 2) 4(1/3 2)

e. 421/3 2

61/3= 7(1/3 2) = 72/3

a. 71/4 71/2 = 7(1/4 + 1/2) = 73/4

= 12[5 (–1/5)]c. (45 35)–1/5 = [(4 3)5]–1/5 = (125)–1/5 112

=

d.

551/3

= 5(1 – 1/3) = 52/351

51/3=

= 426

1/32

EXAMPLE 3 Use properties of radicals

Use the properties of radicals to simplify the expression.

a. 123 183 12 183= 2163= = 6 Product property

b. 804

54

8054= = 164 = 2 Quotient property

EXAMPLE 4 Write radicals in simplest form

Write the expression in simplest form.

a. 1353 = 273 5

= 273 53

533=

Factor out perfect cube.

Product property

Simplify.

EXAMPLE 4 Write radicals in simplest form

285

2= Simplify.

b. 75

85

75

85

45

45= Make denominator a perfect fifth

power.

325

285= Product property

EXAMPLE 5 Add and subtract like radicals and roots

Simplify the expression.

a. 104 1047+ = 104(1 + 7) = 1048

b. (81/5)2 + (81/5)10 = (81/5)(2 +10) = (81/5)12

233 23–=c. 543 – 23 = 23273 23– 23(3 – 1)= = 2 23

GUIDED PRACTICE for Examples 3, 4, and 5

Simplify the expression.

3SOLUTION

6. 274 34

SOLUTION

7. 23

2503

5

245

2SOLUTION

345

8.

SOLUTION

9. 53 403+

3 53

EXAMPLE 6 Simplify expressions involving variables

Simplify the expression. Assume all variables are positive.

a. 64y63 = 43(y2)33 433 (y2)33= = 4y2

b. (27p3q12)1/3 = 271/3(p3)1/3(q12)1/3 = 3p(3 1/3)q(12 1/3) = 3pq4

c. m4

n8

4 =m44

n84 =

m44

(n2)44=

mn2

d.14xy 1/3

2x 3/4 z –6= 7x(1 – 3/4)y1/3z –(–6) = 7x1/4y1/3z6

EXAMPLE 7 Write variable expressions in simplest form

Write the expression in simplest form. Assume all variables are positive.

a. 4a8b14c55 = 4a5a3b10b4c55

a5b10c55 4a3b45=

4a3b45ab2c=

Factor out perfect fifth powers.Product property

Simplify.

b. xy8

3 =x

y8

y

y3

=x yy9

3

Make denominator a perfect cube.

Simplify.

EXAMPLE 7 Write variable expressions in simplest form

x y3

y93=

x y3

y3=

Quotient property

Simplify.

EXAMPLE 8

Perform the indicated operation. Assume all variables are positive.

a. 15 w +

35 w = 4

5 w= 15 + 3

5 w

b. 3xy1/4 8xy1/4– = (3 – 8) xy1/4 = –5xy1/4

12c. z2z53 – 54z23 = 12z 2z23 – 3z 2z23

(12z – 3z) 2z23=

9z 2z23=

Add and subtract expressions involving variables


Simplify the expression. Assume all variables are positive.

3q3

10. 27q93

SOLUTION

SOLUTION

11. x10

y5

5

x2

y

SOLUTION

SOLUTION

2x1/2y1/4

12. 3x 1/2 y 1/2

6xy 3/4

13. w9w5 – w3

2w2 w

ESSENTIAL QUESTION

HOW ARE THE PROPERTIES OF

RATIONAL EXPONENTS RELATED TO PROPERTIES

OF INTEGER EXPONENTS?

All properties of integer exponents also apply to rational exponents

LET F(X) = 3X + 5. FIND F(-6)

LESSON 3 PERFORM FUNCTION OPERATIONS AND

COMPOSITION

ESSENTIAL QUESTION

WHAT OPERATIONS CAN BE PERFORMED ON A

PAIR OF FUNCTIONS TO OBTAIN A THIRD

FUNCTION?

VOCABULARY

• Power Function: A function of the form y=axb, where a is a real number and b is a rational number

• Composition: The composition of a function g with a function f is h(x) = f(f(x)).

SOLUTION

EXAMPLE 1 Add and subtract functions

Let f (x) = 4x1/2 and g(x) = –9x1/2. Find the following.

a. f(x) + g(x)

f (x) + g(x) = 4x1/2 + (–9x1/2) = [4 + (–9)]x1/2 = –5x1/2

b. f(x) – g(x)

SOLUTION

f (x) – g(x) = [4 – (–9)]x1/2 = 13x1/2= 4x1/2 – (–9x1/2)

SOLUTION

EXAMPLE 1 Add and subtract functions

c. the domains of f + g and f – g

The functions f and g each have the same domain: all nonnegative real numbers. So, the domains of f + g and f – g also consist of all nonnegative real numbers.

SOLUTION

EXAMPLE 2 Multiply and divide functions

Let f (x) = 6x and g(x) = x3/4. Find the following.

a. f (x) g(x)

f (x) g(x)

f (x)g(x) =

6xx3/4 = 6x(1 – 3/4) = 6x1/4

= (6x)(x3/4) = 6x(1 + 3/4) = 6x7/4

b. f (x)g(x)

SOLUTION

SOLUTION

EXAMPLE 2 Multiply and divide functions

c. the domains of f g and fg

The domain of f consists of all real numbers, and the domain of g consists of all nonnegative real numbers. So, the domain of f g consists of all nonnegative real numbers. Because g(0) = 0, the domain of is restricted to all positive real numbers.

fg

Rhinos

EXAMPLE 3 Solve a multi-step problem

For a white rhino, heart rate r (in beats per minute) and life span s (in minutes) are related to body mass m (in kilograms) by these functions:

r(m) = 241m–0.25 s(m) = (6 106)m0.2

• Find r(m) s(m).

• Explain what this product represents.

SOLUTION


STEP 1

Find and simplify r(m) s(m).

= 241(6 106)m(–0.25 + 0.2) Product of powers property

= (1.446 109)m –0.05

Write product of r(m) and s(m).

= 241m –0.25 [ (6 106)m0.2 ]

(1446 106)m –0.05= Simplify.

r(m) s(m)

Use scientific notation.


STEP 2

Interpret r(m) s(m).

Multiplying heart rate by life span gives the total number of heartbeats for a white rhino over its entire lifetime.


Let f (x) = –2x2/3 and g(x) = 7x2/3. Find the following.

f (x) + g(x)1.

SOLUTION

f (x) + g(x) = –2x2/3 + 7x2/3 = (–2 + 7)x2/3 = 5x2/3

f (x) – g(x)2.

SOLUTION

f (x) – g(x) = –2x2/3 – 7x2/3 = [–2 + ( –7)]x2/3 = –9x2/3


SOLUTION

3. the domains of f + g and f – g

all real numbers; all real numbers


Let f (x) = 3x and g(x) = x1/5. Find the following.

SOLUTION

SOLUTION

f (x) g(x)4.

3x6/5

f (x)5.

g(x)

3x4/5


6. f

gthe domains of f g and

all real numbers; all real numbers except x=0.

SOLUTION


Rhinos

7. Use the result of Example 3 to find a white rhino’s number of heartbeats over its lifetime if its body mass is 1.7 105 kilograms.

about 7.92 108 heartbeatsSOLUTION

ESSENTIAL QUESTION

WHAT OPERATIONS CAN BE PERFORMED ON A

PAIR OF FUNCTIONS TO OBTAIN A THIRD

FUNCTION?Two functions can be combined by the operations:

+, -, x, ÷ and composition

SOLVE X=4Y3 FOR Y

LESSON 4: USE INVERSE FUNCTIONS

ESSENTIAL QUESTION

HOW DO YOU FIND AN INVERSE RELATION OF A

GIVEN FUNCTION?

VOCABULARY

• Inverse relation: A relation that interchanges the input and output values of the original relation. The graph of an inverse relation is a reflection of the graph of the original relation, with y=x as the line of reflection

• Inverse function: An inverse relation that is a function. Functions f and g are inverses provided that f(g(x)) = x and g(f(x)) = x

EXAMPLE 1 Find an inverse relation

Find an equation for the inverse of the relation y = 3x – 5.

Write original relation.y = 3x – 5

Switch x and y.x = 3y – 5

Add 5 to each side.x + 5 = 3y

Solve for y. This is the inverse relation.

13 x + 5

3 = y

EXAMPLE 2 Verify that functions are inverses

Verify that f(x) = 3x – 5 and f –1(x) = 13 x + 5

3are inverse functions.

STEP 1Show: that f(f –1(x)) = x.

f (f –1(x)) = f 31 x + 5

3

= x + 5 – 5

= x

SOLUTION

31 x + 5

3= 3 – 5

STEP 2Show: that f –1(f(x)) = x.

= 13

53(3x – 5) +

= x – 53

53+

= x

f –1(f(x)) = f –1((3x – 5)

for Examples 1, 2, and 3GUIDED PRACTICE

Find the inverse of the given function. Then verify that your result and the original function are inverses.

1. f(x) = x + 4

x – 4 = y

2. f(x) = 2x – 1

x + 12 = y

3. f(x) = –3x – 1

x 13 = ySOLUTION

SOLUTION

SOLUTION

GUIDED PRACTICE

4. Fitness: Use the inverse function in Example 3 to find the length at which the band provides 13 pounds of resistance.

SOLUTION 48 inches

for Examples 1, 2, and 3

ESSENTIAL QUESTION

HOW DO YOU FIND AN INVERSE RELATION OF A

GIVEN FUNCTION?Write the original equation.

Switch x and y. Solve for y.

EXPAND AND SOLVE:(X-5)2

LESSON 6: SOLVE RADICAL EQUATIONS

ESSENTIAL QUESTION

WHY IS IT NECESSARY TO CHECK EVERY APPARENT SOLUTION OF A RADICAL

EQUATION IN THE ORIGINAL EQUATION?

VOCABULARY

• Radical equation: An equation with one or more radicals that have variables in their radicands

• Extraneous solution: An apparent solution that must be rejects because it does not satisfy the original equation.

EXAMPLE 1 Solve a radical equation

Solve 3 = 3. 2x+7

2x+73 = 3

= 33 2x+73( )3

2x+7 = 27

2x = 20

x = 10

Write original equation.

Cube each side to eliminate the radical.

Simplify.

Subtract 7 from each side.

Divide each side by 2.

EXAMPLE 1 Solve a radical equation

CHECK

Check x = 10 in the original equation.

Substitute 10 for x.

Simplify.

Solution checks.= 33

2(10)+73 3=?

273 3=?

GUIDED PRACTICE for Example 1

Solve equation. Check your solution.

1. 3√ x – 9 = –1

x = 512

x = –9

2. ( x+25 ) = 4

x = 11

3. (23 x – 3 ) = 4

ANSWER

ANSWER

ANSWER

ESSENTIAL QUESTION

WHY IS IT NECESSARY TO CHECK EVERY APPARENT SOLUTION OF A RADICAL

EQUATION IN THE ORIGINAL EQUATION?

Raising both sides of an equation to the same power sometimes results in an extraneous

solution

please complete the prerequisite skills pg 412 #1-12

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