Please complete the prerequisite Skills PG 412 #1-12
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Transcript of 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
GUIDED PRACTICE for Examples 1, 2 and 3
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
GUIDED PRACTICE for Examples 1, 2 and 3
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
GUIDED PRACTICE for Examples 6, 7, and 8
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
EXAMPLE 3 Solve a multi-step problem
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.
EXAMPLE 3 Solve a multi-step problem
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.
GUIDED PRACTICE for Examples 1, 2, and 3
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
GUIDED PRACTICE for Examples 1, 2, and 3
SOLUTION
3. the domains of f + g and f – g
all real numbers; all real numbers
GUIDED PRACTICE for Examples 1, 2, and 3
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
GUIDED PRACTICE for Examples 1, 2, and 3
6. f
gthe domains of f g and
all real numbers; all real numbers except x=0.
SOLUTION
GUIDED PRACTICE for Examples 1, 2, and 3
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