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Please complete the prerequisite Skills PG 412 #1-12. Chapter 6: Rational Exponents and Radical Functions. Big ideas: Use Rational Exponents Performing function operations and finding inverse functions Solving radical equations. Lesson 1: Evaluate nth Roots and Use Rational Exponents.

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Please complete the prerequisite SkillsPG 412 #1-12


Chapter 6:Rational Exponents and Radical Functions

Big ideas:

Use Rational Exponents

Performing function operations and finding inverse functions

Solving radical equations


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


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.

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

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


1

1

23

323/5

64

( )3

=

(161/2)3

=

43

=

43

=

64

=

=

16

1

1

1

1

1

1

=

=

=

=

( )3

(321/5)3

323/5

32

5

8

23

8

=

=

=

=

EXAMPLE 2

Evaluate expressions with rational exponents

Evaluate (a) 163/2 and (b)32–3/5.

SOLUTION

Radical Form

Rational Exponent Form

a. 163/2

163/2

b. 32–3/5

32–3/5


Keystrokes

Expression

Display

9 1 5

7 3 4

12 3 8

7

c. ( 4 )3 = 73/4

EXAMPLE 3

Approximate roots with a calculator

a. 91/5

1.551845574

b. 123/8

2.539176951

4.303517071


for Examples 1, 2 and 3

GUIDED PRACTICE

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

1. n = 4, a = 625

3. n = 3, a = –64.

SOLUTION

±5

SOLUTION

–4

2.n = 6, a = 64

4. n = 5, a = 243

SOLUTION

±2

SOLUTION

3


1

3

for Examples 1, 2 and 3

GUIDED PRACTICE

Evaluate expressions without using a calculator.

5. 45/2

7. 813/4

27

SOLUTION

32

SOLUTION

6. 9–1/2

8. 17/8

SOLUTION

SOLUTION

1


Expression

10. 64 2/3

11. (4√ 16)5

12. (3√–30)2

for Examples 1, 2 and 3

GUIDED PRACTICE

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

9. 42/5

1.74

SOLUTION

SOLUTION

0.06

SOLUTION

32

9.65

SOLUTION


a. 4x5

= 128

x5

=

32

x

=

32

5

x

2

=

EXAMPLE 4

Solve equations using nth roots

Solve the equation.

Divide each side by 4.

Take fifth root of each side.

Simplify.


b. (x – 3)4

= 21

+

x – 3

=

21

4

+

x

=

21

+ 3

4

or

21

+ 3

x

=

x

=

21

+ 3

4

4

5.14

0.86

x

or

x

EXAMPLE 4

Solve equations using nth roots

Take fourth roots of each side.

Add 3 to each side.

Write solutions separately.

Use a calculator.


Essential question

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

What is the relationship between nth roots and rational exponents?


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


51

5

51/3

51/3

a. 71/4 71/2

b. (61/2 41/3)2

= (61/2)2 (41/3)2

= 6(1/22) 4(1/32)

= 6 42/3

= 61 42/3

1

c. (45 35)–1/5

= [(4 3)5]–1/5

= 12[5 (–1/5)]

=

12

d.

=

2

42

421/3 2

1/3

e.

=

= 7(1/3 2)

6

61/3

EXAMPLE 1

Use properties of exponents

Use the properties of rational exponents to simplify the expression.

= 73/4

= 7(1/4 + 1/2)

= 12 –1

= (125)–1/5

= 5(1 – 1/3)

= 52/3

= 72/3

= (71/3)2


a.

5

80

12

16

18

216

=

12 18

=

=

6

4

4

4

3

3

3

3

80

b.

=

=

=

2

4

5

EXAMPLE 3

Use properties of radicals

Use the properties of radicals to simplify the expression.

Product property

Quotient property


5

5

3

3

3

3

27

5

27

a.

=

=

3

=

3

135

EXAMPLE 4

Write radicals in simplest form

Write the expression in simplest form.

Factor out perfect cube.

Product property

Simplify.


7

7

8

4

8

4

5

5

5

5

5

5

5

5

5

28

28

32

=

b.

=

=

2

EXAMPLE 4

Write radicals in simplest form

Make denominator a perfect fifth power.

Product property

Simplify.


2

2

2

2

2

3

3

3

3

3

(1 + 7)

a.

8

7

+

=

=

4

3

4

4

4

3

3

10

10

54

27

10

10

b.

=

=

+

(81/5)

2

3

(3 – 1)

c.

2

3

=

=

=

2

(81/5)

(81/5)

12

10

2

=

(2 +10)

(81/5)

EXAMPLE 5

Add and subtract like radicals and roots

Simplify the expression.


3

5

3

24

2

3

250

+

40

3

5

4

4

4

27

3

3

3

3

5

5

3

5

2

for Examples 3, 4, and 5

GUIDED PRACTICE

Simplify the expression.

SOLUTION

SOLUTION

SOLUTION

SOLUTION


3

43(y2)3

a.

4y2

=

=

=

3pq4

b.

(27p3q12)1/3

271/3(p3)1/3(q12)1/3

=

=

=

3

4

4

4

3p(3 1/3)q(12 1/3)

n8

43

m4

m4

m

3

3

(y2)3

64y6

14xy 1/3

4

7x1/4y1/3z6

7x(1 – 3/4)y1/3z –(–6)

=

=

c.

=

=

=

n2

2x 3/4 z –6

4

(n2)4

d.

m4

n8

EXAMPLE 6

Simplify expressions involving variables

Simplify the expression. Assume all variables are positive.


5

=

5

4a8b14c5

4a5a3b10b4c5

5

5

a5b10c5

4a3b4

a.

=

=

b.

=

x

5

ab2c

4a3b4

y8

x y

3

x

y

3

=

3

y9

y8

y

EXAMPLE 7

Write variable expressions in simplest form

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

Factor out perfect fifth powers.

Product property

Simplify.

Make denominator a perfect cube.

Simplify.


3

xy

=

3

x y

3

=

y9

y3

EXAMPLE 7

Write variable expressions in simplest form

Quotient property

Simplify.


3z)

(12z

1

3

w

w

+

+

5

5

a.

=

=

9z

3

3

2z2

2z5

3

1

4

(3 – 8) xy1/4

–5xy1/4

b.

=

=

3xy1/4

8xy1/4

5

5

5

c.

=

z

=

12

w

w

=

3

3

3

3

3z

2z2

2z2

54z2

2z2

12z

EXAMPLE 8

Add and subtract expressions involving variables

Perform the indicated operation. Assume all variables are positive.


6xy 3/4

3x 1/2 y 1/2

3q3

3

27q9

5

2x1/2y1/4

w

w3

9w5

x10

y5

x2

y

w

2w2

for Examples 6, 7, and 8

GUIDED PRACTICE

Simplify the expression. Assume all variables are positive.

SOLUTION

SOLUTION

SOLUTION

SOLUTION


Essential question

All properties of integer exponents also apply to rational exponents

How are the properties of rational exponents related to properties of integer exponents?


Let f(x) = 3x + 5. Find f(-6)


Lesson 3Perform 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)).


a.

f(x) + g(x)

f(x) – g(x)

b.

EXAMPLE 1

Add and subtract functions

Letf (x)= 4x1/2andg(x)=–9x1/2. Find the following.

SOLUTION

f (x) + g(x)

= 4x1/2 + (–9x1/2)

= [4 + (–9)]x1/2

= –5x1/2

SOLUTION

f (x) – g(x)

= [4 – (–9)]x1/2

= 13x1/2

= 4x1/2 – (–9x1/2)


The functions fand geach have the same domain: all nonnegative real numbers. So, the domains of f + gand f – galso consist of all nonnegative real numbers.

c.

the domains of f + gand f – g

EXAMPLE 1

Add and subtract functions

SOLUTION


a.

f (x) g(x)

b.

6x

f (x)

f (x)

g(x)

g(x)

x3/4

f (x) g(x)

=

=

6x1/4

=

6x(1 – 3/4)

EXAMPLE 2

Multiply and divide functions

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

SOLUTION

= (6x)(x3/4)

= 6x(1 + 3/4)

= 6x7/4

SOLUTION


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

f

g

EXAMPLE 2

Multiply and divide functions

f

g

and

the domains of f

c.

g

SOLUTION


r(m)

s(m)

(6 106)m0.2

=

241m–0.25

=

• Findr(m) s(m).

EXAMPLE 3

Solve a multi-step problem

Rhinos

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:

• Explain what this product represents.


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

r(m) s(m)

=

241m –0.25 [ (6 106)m0.2 ]

241(6 106)m(–0.25 + 0.2)

=

(1446 106)m –0.05

=

(1.446 109)m –0.05

=

EXAMPLE 3

Solve a multi-step problem

SOLUTION

STEP 1

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

Product of powers property

Simplify.

Use scientific notation.


Interpret r(m) s(m).

EXAMPLE 3

Solve a multi-step problem

STEP 2

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


f (x) + g(x)

f (x) – g(x)

for Examples 1, 2, and 3

GUIDED PRACTICE

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

SOLUTION

f (x) + g(x)

= –2x2/3 + 7x2/3

= 5x2/3

= (–2 + 7)x2/3

SOLUTION

f (x) – g(x)

= –2x2/3 – 7x2/3

= [–2 + ( –7)]x2/3

= –9x2/3


the domains of f + gand f – g

for Examples 1, 2, and 3

GUIDED PRACTICE

SOLUTION

all real numbers; all real numbers


f (x) g(x)

f (x)

g(x)

for Examples 1, 2, and 3

GUIDED PRACTICE

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

SOLUTION

3x6/5

SOLUTION

3x4/5


the domains off g and

f

g

for Examples 1, 2, and 3

GUIDED PRACTICE

SOLUTION

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


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 105kilograms.

about 7.92 108 heartbeats

for Examples 1, 2, and 3

GUIDED PRACTICE

Rhinos

SOLUTION


Essential question

Two functions can be combined by the operations: +, -, x, ÷ and composition

What operations can be performed on a pair of functions to obtain a third function?


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


5

3

x+

=y

1

3

EXAMPLE 1

Find an inverse relation

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

y = 3x – 5

Write original relation.

x = 3y – 5

Switch x and y.

x + 5 = 3y

Add 5 to each side.

Solve for y. This is the inverse relation.


x

+

Verify thatf(x) = 3x – 5 and f –1(x) =

5

5

1

1

5

5

are inverse functions.

3

3

3

3

3

3

= x

1

1

5

5

x +

x +

f (f –1(x)) =f

f –1(f(x)) =

f –1((3x – 5)

3

3

3

3

(3x – 5) +

=

– 5

= 3

= x –

+

= x

EXAMPLE 2

Verify that functions are inverses

SOLUTION

STEP 1

STEP 2

Show: that f(f –1(x)) = x.

Show: that f –1(f(x)) = x.

= x + 5 – 5


x 1

x + 1

=y

=y

3

2

for Examples 1, 2, and 3

GUIDED PRACTICE

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

1. f(x) = x + 4

3. f(x) = –3x – 1

SOLUTION

x – 4 = y

SOLUTION

2. f(x) = 2x – 1

SOLUTION


for Examples 1, 2, and 3

GUIDED PRACTICE

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

48 inches

SOLUTION


Essential question

Write the original equation.

Switch x and y.

Solve for y.

How do you find an inverse relation of a given function?


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.


2x+7

2x+7

2x+7

Solve 3 = 3.

= 3

3

( )3

3

= 33

2x+7

= 27

2x

= 20

x

= 10

EXAMPLE 1

Solve a radical equation

Write original equation.

Cube each side to eliminate the radical.

Simplify.

Subtract 7 from each side.

Divide each side by 2.


3

27

2(10)+7

3

3

3

= 3

3

?

?

=

=

EXAMPLE 1

Solve a radical equation

CHECK

Check x = 10 in the original equation.

Substitute 10 for x.

Simplify.

Solution checks.


2. ( x+25 ) = 4

3. (23 x –3 ) = 4

for Example 1

GUIDED PRACTICE

Solve equation. Check your solution.

1.3√ x – 9 = –1

x = 512

ANSWER

ANSWER

x = 11

x = –9

ANSWER


Essential question

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

Why is it necessary to check every apparent solution of a radical equation in the original equation?


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