Document related concepts
no text concepts found
Transcript
```Section 10.1
Expressions and
Functions
Objectives
•
•
The Square Root Function
•
The Cube Root Function
Every positive number a has two square roots, one
positive and one negative. Recall that the positive square
root is called the principal square root.
The symbol
The expression under the radical sign is called the
6  x  2, and
5x
3x  4
Example
Evaluate each square root.
a.
36  6
b.
0.64  0.8
c.
4
16

5
25
Example
Approximate 38 to the nearest thousandth.
Solution
 6.164
Example
Evaluate the cube root.
a.
3
64  4
b.
3
125  5
c.
3
1
8
1

2
Example
Find each root, if possible.
a.
4
256
b.
5
243
c. 4 1296
Solution
a.
b.
c.
4
5
4
256  4 because 4  4  4  4  256.
243
1296
 3 because (  3)5  243.
An even root of a negative number is
not a real number.
Example
Write each expression in terms of an absolute value.
a.
( 5) 2
b.
( x  3) 2
c.
Solution
a. ( 5) 2  5  5
b.
c.
( x  3) 2
 x3
w2  6w  9  ( w  3) 2
 w3
w2  6w  9
Example
If possible, evaluate f(1) and f(2) for each f(x).
2
f
(
x
)

x
4
f
(
x
)

5
x

1
a.
b.
Solution
a. f ( x )  5 x  1
f (1)  5(1)  1
 6
f ( 2)  5( 2)  1
 9  undefined
b.
f (x)  x2  4
f (1)  12  4
 5
f ( 2)  ( 2)2  4
 8
Example
Calculate the hang time for a ball that is kicked 75 feet
into the air. Does the hang time double when a ball is
kicked twice as high? Use the formula
1
x
2
Solution
1
The hang time is T (75)  75  4.3 sec
2
T (x) 
The hang time is T (150) 
1
150  6.1 sec
2
The hang times is less than double.
Example
interval notation.
a. f ( x )  3  4 x
b. f ( x )  x 2  4
Solution
Solve 3 – 4x  0.
3  4x  0
4 x  3
3
x
4
3

The domain is  , 4  .
b. Regardless of the value of
x; the expression is always
positive. The function is
defined for all real
numbers, and it domain is
 ,   .
Section 10.2
Rational
Exponents
Objectives
•
Basic Concepts
•
Properties of Rational Exponents
Example
Write each expression in radical notation. Then evaluate
the expression and round to the nearest hundredth when
appropriate.
a. 49
1/2
b. 26
Solution
a. 491/2  49  7
c. (6 x)1/2  6x
1/5
c.  6x 
1/2
b. 261/5
Example
Write each expression in radical notation. Evaluate the
expression by hand when possible.
2/3
a.  8 
b. 103/4
Solution
2
2/3
3
a.  8    8    2 2  4
b. 103/4  4 103  4 1000
Example
Write each expression in radical notation. Evaluate the
expression by hand when possible.
a. 813/4
b. 144/5
Solution
a. 813/4
Take the fourth root of 81
4/5 Take the fifth root of 14
14
b.
3/ 4 and then cube it.
and then fourth it.
 (81)


4
81
3
3
 27

3
 144/5


5
14

4
Cannot be evaluated by
hand.
Example
Write each expression in radical notation and then
evaluate.
a. 811/4
b. 64 2/3
Solution
1
a. 811/4  1/4
81
1
 4
81
1

3
b. 64 2/3  12/3
64


1
3
64

2
1
1
 2 
4
16
Example
Use rational exponents to write each radical expression.
a. 7 x3  x 3/7
1
b.
b3
 b 3/ 2
c.
5
( x  1) 2
 ( x  1)2/5
d.
4
a b
 (a 2  b2 )1/4
2
2
Example
Write each expression using rational exponents and
simplify. Write the answer with a positive exponent.
Assume that all variables are positive numbers.
a.
x4 x
x
1/2
x
1/4
b.
4
256x3  (256 x3 )1/4
 x1/21/4
 2561/4 ( x3 )1/4
 x 3/4
 4x 3/4
Example (cont)
Write each expression using rational exponents and
simplify. Write the answer with a positive exponent.
Assume that all variables are positive numbers.
5
c.
32x
4
x
(32 x)1/5

x1/4
321/5 x1/5

x1/4
 2x1/51/4
 2x 1/20
2
 1/20
x
x 


27


3
d.
1/3
1/3
 27 
 3 
x 
271/3
 3 1/3
(x )
3

x
Example
Write each expression with positive rational exponents
and simplify, if possible.
1/4
y
a. 4 x  2
b.
1/5
x
Solution
a. 4 x  2  ( x  2)1/2 1/4


 ( x  2)1/8
y 1/4
b. 1/5
x
x1/5
 1/4
y
Section 10.3
Simplifying
Expressions
Objectives
•
•
Consider the following example:
4  25  2  5  10
4  25  100  10
Note: the product rule only works when the radicals have
the same index.
Example
a. 36  4
b.
c.
3
4
8  3 27
 36  4  144  12
 3 8  27  3 216  6
1 4 1 41



4 16 4
4
1 1 1
  
4 16 4
4
1
1

256 4
Example
x 2  x 4  x 2  x 4  x 6  x3
a.
b. 5a  10a
c.
3
3
4
3x 4 7 y

y
x
2
 3 5a 10a2  3 50a3  a 3 50

4
3x 7 y


y x
4
21xy 4
 21
xy
Example
Simplify each expression.
a. 500  100  5  10 5
b.
c.
3
40  3 8  3 5  2 3 5
72  36  2  6 2
Example
Simplify each expression. Assume that all variables are
positive.
5
4
75
y
4
4
2

25y
   3y
a. 49x  49  x  7 x b.
 25y 4  3 y
c.
3
3a  3 9a2 w
 5 y 2 3y
 3 3a  9a2 w


3
3
 27a  w
 27a  
 3a 3 w
3
3
3
w
Example
Simplify each expression.
a.
7  7  7 7
3
1/2
1/3
b.
3
a a
5
 a1/3  a1/5
7
 a1/31/5
 7 5/6
 a 8/15
1/21/3
Quotient Rule
Consider the following examples of dividing radical
expressions:
4
2 2 2
9

4

9

3 3

4 2

9 3
3
Example
Simplify each radical expression. Assume that all
variables are positive.
a.
3
7

27
3
3
3
7
27
7

3
b.
5
x

32
5
x
5
32
5
x

2
Example
Simplify each radical expression. Assume that all
variables are positive.
a.
90
90

10
10
 9
3
b.
x4 y
y

x4 y
y
 x4
 x2
Example
Simplify the radical expression. Assume that all variables
are positive.
5
32x 4
y5

5
32x 4
5

5
y5
32  5 x 4
5
y5
2 5 x4

y
Example
Simplify the expression.
x 1  x  1
Solution
x  1  x  1  ( x 1)( x  1)

x2  1
Example
Simplify the expression.
3
x2  5x  6
3
x2
Solution
3
x2  5x  6

3
x2

3
3
( x  3)( x  2)
( x  2)
x3
Section 10.4
Operations on
Expressions
Objectives
•
•
Multiplication
•
Rationalizing the Denominator
Like radicals have the same index and the same
Like
Unlike
3 2 5 2
3 2 5 3
Example
If possible, add the expressions and simplify.
a. 4 7  8 7  12 7
b. 7 3 5  2 3 5  9 3 5
c. 8  13
d.
6  16
The terms cannot be added because they are not like
The expression contains unlike radicals and cannot be
Example
Write each pair of terms as like radicals, if possible.
3
3
4
16,
7
54
a. 80, 125
b.
Solution
a.
80  16  5  4 5
125  25  5  5 5
b. 4 3 16  4 3 8  3 2
 4  23 2  83 2
7 3 54  7 3 27  3 2
 7  3  3 2  213 2
Example
a. 20  5 5
Solution
a. 20  5 5
b. 5 2  50  72
b. 5 2  50  72
 45  5 5
 5 2  25  2  36  2
 2 5 5 5
=5 2  5 2  6 2
7 5
 16 2
Example
Simplify the expressions.
a. 8 7  2 7  6 7
b. 7 3 5  2 3 5  3 5
 (7  2  1) 3 5  6 3 5
Example
Subtract and simplify. Assume that all variables are
positive.
3 7y
3 7y
3 7y
7
y
a. 49x5  x5
b. 3 64  4  4  4
 49x4  x  x4  x
 7x 2 x  x 2 x
 6x 2 x
0
Example
Subtract and simplify. Assume that all variables are
positive.
a. 7 2  3 2
b. 3 343a7b4  24 3 27ab
5
3
 343a6b3  3 ab  24 3 27  3 ab
3
7 2 3 3 2 5

 

5 3
3 5
 7a 2 b 3 ab  24  3 3 ab
21 2 15 2


15
15
 7a 2 b 3 ab  72 3 ab
6 2 2 2


15
5
 (7a 2 b  72) 3 ab
Example
Multiply and simplify.
3  x 5  x 
Solution
3  x 5  x   3  5  3
x 5 x  x  x
 15  3 x  5 x  x2
 15  2 x  x
Example
Rationalize each denominator. Assume that all variables
are positive.
ab
7
1
a.
b.
c.
5
8 3
3
b
Solution
a. 1  3  3
3
3
3
7
3
7 3
7 3

b.


8 3
3
83
24
c.
ab b
a b
ab  ab
ab
b
 2
 2
 2

2
5
b
b
b
b b
b b b
b
Conjugates
Example
1
Rationalize the denominator of 1  3 .
Solution
1
1
1 3


1 3 1 3 1 3
1 3
 2
1  ( 3) 2
1 3
1 3
1 3

2

1
3

2  2
1
3
 
2
2

Example
Rationalize the denominator of 4  6
3 6
Solution
4 6 3 6
4 6


3 6 3 6
3 6
12  4 6  3 6  ( 6) 2

9  ( 6) 2
18  7 6
3
18 7 6


3
3

7 6
 6
3
Example
Rationalize the denominator of
Solution
4
3
x2
4
 2/3
x
4 x1/3
 2/3  1/3
x
x
4x1/3
 2/31/3
x
43 x

x
4
3
x2
Section 10.6
Equations
Expressions
Objectives
•
•
The Distance Formula
•
Solving the Equation xn = k
POWER RULE FOR SOLVING EQUATIONS
If each side of an equation is raised to the same
positive integer power, then any solutions to the
given equations are among the solutions to the
new equation. That is, the solutions to the
equation a = b are among the solutions to an = bn.
Example
Solve 4 2 x  1  12. Check your solution.
Solution
4 2 x  1  12

2x 1  3
2x 1

2
 32
2x 1  9
2x  8
x4
Check:
4 2  4   1  12
4 9  12
4  3  12
12  12
It checks.
Step 1: Isolate a radical term on one side of
the equation.
Step 2: Apply the power rule by raising each side of
the equation to the power equal to the index
Step 3: Solve the equation. If it still contains radical,
repeat Steps 1 and 2.
result in the given equation.
Example
Solve 6  x  3  1.
Solution
Step 1: To isolate the radical term, we add 3 to each side
of the equation.
6  x 3 1
Step 2: Square each side.
Step 3: Solve the resulting equation.
6 x  4

6 x

2
 42
6  x  16
 x  10
x  10
Example (cont)
6  x  3  1.
equation.
6  x 3 1
6   10  3  1
16  3  1
4 3 1
11
Since this checks, the solution is x = −10.
Example
Solve 3x  2  x  2. Check your results and then solve
the equation graphically.
Solution
Check:
3x  2  x  2
Symbolic Solution

3  6  2   6  2
3x  2  x  2
3x  2

2
  x  2
2
3x  2  x 2  4 x  4
0  x2  7 x  6
0   x  6 x  1
x  6 or x = 1
44
It checks.
3x  2  x  2
3 1  2  1  2
1  1
Thus 1 is an extraneous solution.
Example (cont)
3x  2  x  2
Graphical Solution
The solution 6 is supported graphically where the
intersection is at (6, 4). The graphical solution does not
give an extraneous solution.
Example
Solve 4 x  3  x  3
Solution
4x  3  x  3

4x

2
2


4x  3

 
2
9 x 2  72x  144  144 x
x 3

4 x (3)  32  x  3
3 x  12  6 4 x
 3 x  12

 6 4x
9 x 2  72x  144  0
(3 x  12)2  0
3x  12  0
3 x  12
x4
4x  6 4x  9  x  3
2
2

9 x 2  72 x  144  36(4 x )
2
The solution is 4.
Example
Solve.
3
2x  6  4
Solution
Step 1: The cube root is already isolated, so we proceed
to Step 2.
3
3
3
2
x

6

4


Step 2: Cube each side.
Step 3: Solve the resulting equation.
2 x  6  64
2x  58
x  29
Example (cont)
3
2x  6  4
Step 4: Check the answer by substituting into the given
equation.
3
3
2x  6  4
2  29  6  4
3
64  4
44
Since this checks, the solution is x = 29.
Example
Solve x3/4 = 4 – x2 graphically. This equation would be
difficult to solve symbolically, but an approximate solution
can be found graphically.
Solution
Example
A 6ft ladder is placed against a garage with its base 3 ft
from the building. How high above the ground is the top
Solution
2
2
2
c  a b
62  a 2  32
36  a 2  9
27  a 2
27  a
3 3a
The ladder is 5.2 ft above ground.
DISTANCE FORMULA
The distance d between the points (x1, y1)
and (x2, y2) in the xy-plane is
d
 x2  x1 
2
  y2  y1  .
2
Example
Find the distance between the points (−1, 2) and (6, 4).
Solution
d
 x2  x1    y2  y1  .
d
 6   1   4  2
2
d  49  4
d  53
d  7.28
2
2
2
SOLVING THE EQUATION
xn = k
Take the nth root of each side of xn = k to obtain
n
n
x  n k.
1. If n is odd, then n x n  x and the equation
n
x

k.
becomes
n
x
 x and
2. If n is even and k > 0, then
n
x

k.
the equation becomes
n
(If k < 0, there are no real solutions.)
Example
Solve each equation.
a. x3 = −216
b. x2 = 17
Solution
a. x3  216
b. x 2  17
3
x  216
3
3
x  6
x 2  17
or
x  17
x   17
c. 3(x + 4)4 = 48
Example (cont)
c. 3(x + 4)4 = 48
( x  4)4  16
4
 x  4  4 16
x4  2
x  4  2
x  6
x4 2
x  2
Example
The formula for the volume (V) of a sphere with a radius
4 3
(r), is given by V   r . Solve for r.
3
Solution
4 3
V  r
3
3V
 r3
4
3
3V
r
4
Section 10.7
Complex Numbers
Objectives
•
Basic Concepts
•
•
Powers of i
•
Complex Conjugates and Division
PROPERTIES OF THE IMAGINARY UNIT i
i  1 and
THE EXPRESSION
If a > 0, then
i 2  1
a
a = i a .
Example
Write each square root using the imaginary i.
a.
b. 15
c. 45
36
Solution
a.
36  i 36  6i
b.
15
c.
45  i 45  i 9 5  3i 5
 i 15
SUM OR DIFFERENCE OF COMPLEX NUMBERS
Let a + bi and c + di be two complex numbers. Then
(a + bi) + (c+ di) = (a + c) + (b + d)i
Sum
and
(a + bi) − (c+ di) = (a − c) + (b − d)i. Difference
Example
Write each sum or difference in standard form.
a.
(−8 + 2i) + (5 + 6i)
b. 9i – (3 – 2i)
Solution
a. (−8 + 2i) + (5 + 6i)
= (−8 + 5) + (2 + 6)I
= −3 + 8i
b. 9i – (3 – 2i)
= 9i – 3 + 2i
= – 3 + (9 + 2)I
= – 3 + 11i
Example
Write each product in standard form.
a.
(6 − 3i)(2 + 2i)
b. (6 + 7i)(6 – 7i)
Solution
a.
(6 − 3i)(2 + 2i)
= (6)(2) + (6)(2i) – (2)(3i) – (3i)(2i)
= 12 + 12i – 6i – 6i2
= 12 + 12i – 6i – 6(−1)
= 18 + 6i
Example (cont)
Write each product in standard form.
a.
(6 − 3i)(2 + 2i)
b. (6 + 7i)(6 – 7i)
Solution
b.
(6 + 7i)(6 – 7i)
= (6)(6) − (6)(7i) + (6)(7i) − (7i)(7i)
= 36 − 42i + 42i − 49i2
= 36 − 49i2
= 36 − 49(−1)
= 85
POWERS OF i
The value of in can be found by dividing n (a
positive integer) by 4. If the remainder is r, then
in = ir.
Note that i0 = 1, i1 = i, i2 = −1, and i3 = −i.
Example
Evaluate each expression.
a.
i25
b. i7
c. i44
Solution
a. When 25 is divided by 4, the result is 6 with the
remainder of 1. Thus i25 = i1 = i.
b. When 7 is divided by 4, the result is 1 with the
remainder of 3. Thus i7 = i3 = −i.
c. When 44 is divided by 4, the result is 11 with the
remainder of 0. Thus i44 = i0 = 1.
Example
Write each quotient in standard form.
9
3

2
i
a.
b.
5i
3i
Solution
a. 3  2i   3  2i  5  i   3  5   3  i    2i  5    2i  i 
5i
 5  i  5  i 
5  5   5  i   5  i    i  i 
15  3i  10i  2i 2 15  7i  2  1


2
25   1
25  5i  5i  i
17  7i

26
17 7i


26 26
Example (cont)
b.
9
9  3i 

3i
 3i  3i 
27i

9i 2
27i

9  1
27i

9
 3i
```
Related documents