Name:
1
Final Review
Chapter 1: Graphs, Functions, and Models
1.1
1.1.1
Introduction to Graphing
Know how to graph an equation
Example 1. Create a table of values and graph the equation y = x2 − 1.
Solution
x
−2
f (x) 3
−1
0
0
−1
1
0
2
3
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
1.1.2
The Distance Formula
1.1.3
The Midpoint Formula
1.1.4
Equation of a circle
Example 2. Find the center, radius and equation of the circle that has a diameter with end points
(1, −3) and (3, −1)
Solution
p
√
√
√
d = (3 − 1)2 + (−1 − (−3))2 = 4 + 4 = 2 2 So the radius of the circle is 2. The center of
the
of the line segment.
circle is the midpoint
1 + 3 −3 + (−1)
,
= (2, −2) Using these two facts gives the equation of the circle as (x − 2)2 +
2
2
(y + 2)2 = 2
1.2
1.2.1
Functions and Graphs
Functions and Function Notation:
Example 3. The Domain is grades {A,B,C,D,F} and the Range is Students in a class {Bob, Sue,
Fernando, Olga }. The rule assigns a grade to each student. D → Bob, A → Sue, A → Fernando, and
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A → Olga. Is this relation a function?
Solution
No the input A is sent to 3 different outputs
1.2.2
The Vertical line test
Example 4. Is this a function?
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
Solution
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
No. A vertical line can be drawn that intersects the graph more than one time.
1.2.3
Domain and Range of a function
Example 5. Find the domain of f (x) =
College Algebra
√
x − 3 and g(x) =
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x2 − 1
x+1
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Solution
Set the radicand greater than or equal to zero. x − 3 ≥ 0 ⇐⇒ x ≥ 3 ⇐⇒ [3, ∞)
Set the denominator equal to zero to find the excluded values
x + 1 = 0 ⇐⇒ x = −1 ⇐⇒ {x|x 6= −1}
Example 6. Find the Domain and Range of the following function.
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
Solution
The domain is the setup of inputs that have an out put. We can read the domain from the graph
[−4, 4]. The Range is the set of outputs [0, 3]
1.3
Linear Function, Slope, Applications and models
1.3.1
Horizontal and vertical lines
1.3.2
Slope
Example 7. Find the slope between (−4, 5) and (5, −4).
Solution
m=
1.3.3
−4 − 5
−9
=
= −1
5 − (−4)
9
Equations of a line
Example 8. Find the equation of the line that contains the points (1, 2) and (4, −1).
Solution
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m=
Final Review
−1 − 2
−3
=
= −1 ⇐⇒ 2 = −1(1) + b ⇐⇒ 3 = b ⇐⇒ y = −x + 3
4−1
3
Example 9. Find the equation of the line that contains the point (1, 2) and has slope −3.
Solution
y = −3x + b ⇐⇒ 2 = −3(1) + b ⇐⇒ 5 = b ⇐⇒ y = −3x + 5
1.4
1.4.1
Linear Equations and Modeling
Linear Regression
Example 10. The data in the following table shows the decrease in the number of drive-in movie sites
since 1990.
Year, x Drive in Movie Sites, y
1990, 0
910
1993, 3
837
1996, 6
826
1999, 9
683
2002, 12
666
2005, 15
648
Use your graphing calculator to find a linear regression of the data and use it to estimate the number
of drive-in movie sites in 2015. Find the residual (correlation coefficient) and decide if the line is a good
fit. Please round your answer to the nearest whole number.
Solution
The equation of the line is y = −18.72x + 902.10. The year 2015 corresponds to x = 25.
This gives y = −18.72(25) + 902.10 = 434.1. So in 2015 There will be about 434 Drive in Movie
Sites. r ≈ .95 so the line is a good fit.
Example 11. Is the following data set linear?
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f (x)
6
4
•
•
−6
−4
•
•2
•
−2 •
x
2
4
6
−2
••• • •
•
−4
−6
Solution
No the points do not look like they are on a line.
1.4.2
Parallel and Perpendicular lines
Example 12. Find the equation of the line that is perpendicular to f (x) =
point (−3, −2).
1
x − 2 and contains the
5
Solution
Since perpendicular lines have slopes that are negative reciprocals the slope of the new line is
m = −5. Using the point slope formula gives y = −5x + b ⇐⇒ −2 = −5(−3) + b ⇐⇒ −17 = b
This gives the equation of the line as y = −5x − 17
1.5
1.5.1
Zeros of a function and Applications
Zero’s of linear functions
2
Example 13. Find all zero’s of f (x) = − x − 2.
3
Solution
2
2
To find a zero of a function set y = 0. This gives 0 = − x − 2 ⇐⇒ 2 = − x ⇐⇒ −3 = x
3
3
1.5.2
Application problems
Example 14. Erica invested a total of $5,000 in two accounts. The 1st account pays 3% simple interest per year and the 2nd account pays 4% simple interest per year. After one year she earned $176 in
interest, how much did she invest in each account?
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Solution
Construct a table to organize the information.
I
=
P
r
t
3%
.03x
x
.03
1
4%
.04(5000 − x)
5000 − x
.04
1
total
176
5000
N.A. 1
The interest column gives the equation .03x + .04(5000 − x) = 176 ⇐⇒ x = 2400. So the 3%
account has $2400 and the 4% account has $2600.
1.6
1.6.1
Solving Linear Inequalities
Solve linear inequalities
Example 15. 8x − 4 ≤ 5x + 2
Solution
8x − 4 ≤ 5x + 2
3x ≤ 6
x≤2
x
−5
−4
−3
−2
−1
0
1
2
3
4
5
Example 16. Solve 2x − 1 > 3 or 2x − 1 < −3
Solution
2x − 1 > 3
2x > 4
x>2
2x − 1 < −3
2x < −2
x < −1
x
−5
−4
−3
−2
−1
0
1
2
3
4
5
4
5
Example 17. −5 ≤ 2x − 1 < 7
Solution
−5
−4
−2
≤ 2x − 1 < 7
≤
2x
< 8
≤
x
< 4
x
−5
−4
−3
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−2
−1
0
1
2
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3
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2
Final Review
Chapter 2: More on Functions
Example 18. Determine where the function is increasing decreasing and constant. Write your solution
in interval notation.
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
Solution
The function is increasing on (−3, −1), decreasing on (−∞, −3) ∪ (−1, 1) ∪ (4, ∞) and constant
on (1, 4).
2.0.2
Relative Maximum and Minimum
Example 19. Find the local max and min on the graph below.
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
Solution
The graph has a local max at (−1, 5) and a local min at (−3, 1)
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Example 20. Use a graphing calculator to find the maximum(s) of f (x) = −x4 + 2x2 − 1
Solution
graphing on a TI gives
The max occurs at (±1, 0) and the local min is at (0, −1)
2.1
The Algebra of Functions
2.1.1
Sums, Differences, Products and Quotients
√
2
f
Example 21. If f (x) = x + 1 and g(x) = , find
(x) and state its domain.
x
g
Solution
The domain of f (x) is [−1, ∞) and to domain of g(x) is (−∞, 0) ∪ (0, ∞). Since g(x) is never
equal to zero we only need to find the intersection of the domains. This gives [−1, 0) ∪ (0, ∞)
2.1.2
Difference Quotient
f (x + h) − f (x)
h
Example 22. Find the difference quotient of the function f (x) = 2x + 1
Solution
2(x + h) + 1 − (2x + 1)
2h
=
=2
h
h
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2.2
2.2.1
Final Review
The Composition of Functions.
Function Composition
Example 23. If f (x) =
1
and g(x) = x2 − 1 find (f ◦ g)(x) and (f ◦ f )(x) and state the domain.
x
Solution
We must restrict the domain of g(x) so its range is in the domain of f (x). We must exclude the
1
zero(s) of g(x). This gives x 6= ±1. (f ◦ g)(x) = 2
with domain {x|x 6= ±1}
x −1
Since f (x) is never equal to zero we do not need to exclude any extra values from the domain,
1
but the domain of f (x) is {x|x 6= 0}. So (f ◦ f )(x) = 1 = x with domain {x|x 6= 0}
x
2.2.2
Function Decomposition
p
Example 24. Write h(x) = x2 − x − 6 as the composition of two functions.
Solution
√
f (x) = x
2.3
2.3.1
g(x) = x2 − x − 6
(f ◦ g)(x) = h(x)
Symmetry and Transformations
x-axis, y-axis, and the origin
Example 25. Determine the symmetries of the equation:
x2
y2
−
=1
9
4
Solution
Replacing x with −x gives
y2
(−x)2
y2
x2
−
= 1 ⇐⇒
−
=1
9
4
9
4
Since we get the same equation back the graph is symmetric with respect to the y-axis
Replacing y with −y gives
x2
y2
x2
(−y)2
−
= 1 ⇐⇒
−
=1
9
4
9
4
Since we get the same equation back the graph is symmetric with respect to the x-axis.
Since the graph has both x and y axis symmetry it is also symmetric with respect to the origin.
2.3.2
Even, Odd, and Neither
Example 26. Determine if the function f (x) = x3 − x is even, odd or neither.
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Solution
f (−x) = (−x)3 − (−x) = −x3 + x = −(x3 − x) = −f (x) ⇐⇒ f (−x) = −f (x). So the function
is odd.
2.4
Translations of Basic Graphs
Example 27. Graph the function f (x) = (x − 5)2 − 3 and g(x) =
1
2
(x − 1) + 2
2
Solution
Both f and g are translations of the basic parabola y = x2 (dashed graph). f is shifted 5 units
to the right and down three units. g is shifted 1 unit to the right,up two units, and shrinks by a
factor of one half.
f (x)
9
6
3
−9 −6 −3
−3
3
6
9
x
−6
−9
Example
28. If (4, −2) and (1, 5) are on the graph of y = f (x) Find a point on the graph y =
1
2·f
·x +1
4
Solution
1
y =2·f
· (4) + 1 = 2 · f (1) + 1 = 2(5) + 1 = 11
4
Example 29. The graph of f (x) is given below. Use it to graph g(x) = −f (x + 4) + 3
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y
10
8
6
4
2
x
−10
−8
−6
−4
−2
2
4
6
8
10
−2
−4
−6
−8
−10
Solution
The graph has one reflections and two shifts. The negative in front of f is a reflection across the
x axis. The x + 4 composed with f (x) shifts the graph four units to the left and the +3 shifts the
graph up 3 units.
y
10
8
6
4
2
x
−10
−8
−6
−4
−2
2
4
6
8
10
−2
−4
−6
−8
−10
2.5
2.5.1
Variation
Direct Variation
Example 30. y varies directly as x, when x = 1 y = 2 Find the constant of variation and y when x = 10
Solution
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y = kx ⇐⇒ 2 = k(1) ⇐⇒ y = 2x ⇐⇒ y = 2(10) = 20
2.5.2
Inverse Variation
Example 31. y is inversely proportional to x, when x = 3 y = 1 Find the constant of variation and y
when x = 10
Solution
y=
2.5.3
k
3
3
k
⇐⇒ 1 =
⇐⇒ k = 3 ⇐⇒ y =
⇐⇒ y =
x
3
x
10
Joint Variation
Example 32. z varies directly as x and y, when x = 3, y = 1 and z = 6. Find the constant of variation
and z when x = 2 and y = 3.
Solution
z = kxy ⇐⇒ 6 = k(3)(1) ⇐⇒ k = 2 ⇐⇒ z = 2xy ⇐⇒ z = 2(2)(3) = 12
2.5.4
Applications of Variation
Example 33. The gravitational force between two objects is directly proportional to their masses (m1
and m2 ) and is inversely proportional to the square of the distance (r) between them. What is the
equation of variation?
Solution
m1 m2
F =k 2
r
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3
Final Review
Chapter 3: Quadratic Functions and Equations; Inequalities
3.1
The Complex Numbers
3.1.1
The Complex Number System
√
√
√
Example 34. Simplify −7
−16
−48
Solution
√
√
−7 = i 7
3.1.2
√
√
−16 = i 16 = 4i
√
√
√
−48 = i 16 · 3 = (4 3)i
Addition and Subtraction
Example 35. Add (3 − 4i) + (−2 + 4i)
Solution
(3 − 4i) + (−2 + 4i) = (3 + (−2)) + ((−4) + 4)i = 1 + 0i = 1
Example 36. Subtract (4 −
√
−25) − (4 −
√
−64)
Solution
√
√
(4 − −25) − (4 − −64) = (4 − 5i) − (4 − 8i) = (4 − 4) + (−5 − (−8))i = 0 + 3i = 3i
3.1.3
Multiplication
√
√
Example 37. −9 · −64
Solution
√
√
−9 · −64 = (3i)(8i) = 24i2 = −24
Example 38. (5 − 6i)(1 + 2i)
Solution
(5 − 6i)(1 + 2i) = 5 + 10i − 6i − 12i2 = 17 + 4i
3.1.4
Powers of i
Example 39. Simplify i99
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Solution
4 divides into 99 24 times with a remainder of 3. This gives i99 = i3 = −i
3.1.5
Conjugates and Division
Example 40. Find the conjugate of 2 − i
Solution
2+i
Example 41. Find the conjugate of 6
Solution
6
Example 42. Find the conjugate of 4 + 4i
Solution
4 − 4i
Example 43. Find the conjugate of −16i
Solution
+16i
Example 44.
i
2+i
Solution
i
2−i
2i − i2
1 + 2i
1 2
·
=
= + i
= 2
2
2+i
2−i
2 −i
5
5 5
3.2
3.2.1
Quadratic Equations, Functions, Zeros, and Models
Quadratic Equations and Functions
Example 45. Solve x2 − 5x = 6
Solution
First the equation must be put into standard form by subtracting 6 from both sides x2 −5x−6 = 0.
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The quadratic can be factored as (x − 6)(x + 1) = 0. Using the zero product principle gives
x − 6 = 0 =⇒ x = 6 or x + 1 = 0 =⇒ x = −1
Example 46. Solve x2 + 64 = 0
Solution
2
First isolate the square
√ term
√ by subtracting 64 from both sides x = −64. Taking the square root
of both sides gives x2 = −64 =⇒ x = ±8i
3.2.2
Completing the Square
Example 47. Use completing the square and the square root principle to solve x2 + 4x − 11 = 0
Solution
Isolate the x2 and x term by adding 11 to both sides and then add
2
b
to both sides of the
2
equation
x2 + 4x = 11 ⇐⇒ x2 + 4x + 4 = 11 + 4 ⇐⇒ x2 + 4x + 4 = 15
Factor the perfect square trinomial
and solve using √
the square root principle.
√
(x + 2)2 = 15 ⇐⇒ x + 2 = ± 15 ⇐⇒ x = −2 ± 15
Example 48. Use completing the square and the square root principle to solve 3x2 + 2x − 9 = 0
Solution
2
b
Isolate the x and x term by adding 9 to both sides and then divide by 3, and finally add
2
to both sides of the equation
2
2
1
1
2
1
28
3x2 + 2x = 9 ⇐⇒ x2 + x = 3 ⇐⇒ x2 + x + = 3 +
⇐⇒ x2 + x + =
3
3
9
9
3
9
9
Factor the perfect square trinomial and solve using the square root principle.
r
√
√
2
28
1
28
1 2 7
−1 ± 2 7
1
x+
=
=⇒ x + = ±
⇐⇒ x = − ±
⇐⇒ x =
3
9
3
9
3
3
3
2
3.2.3
Using the Quadratic Formula
Example 49. Solve: 3x2 − 7x = −2
Solution
2
3x − 7x + 2 = 0 =⇒ a = 3, b − 7, and c = 2 This gives x =
√
7 ± 25
7+5
7−5
1
x=
=⇒ x =
= 2 or x =
=
6
6
6
3
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Math 151
−(−7) ±
p
(−7)2 − 4(3)(2)
⇐⇒
2(3)
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Name:
3.2.4
Final Review
The Discriminant
Example 50. Determine the number and type of solutions of 49x2 − 70x + 25 = 0.
Solution
b2 − 4ac = (−70)2 − 4(49)(25) = 4900 − 4900 = 0. Since the discriminant is zero the equation has
one repeated real solution.
3.2.5
Equations Reducible to Quadratic
Example 51. Solve (x + 3)2 − 5(x + 3) − 6 = 0
Solution
The equation is quadratic in form. Let u = x + 3 , then u2 − 5u − 6 = 0 Factoring gives
(u − 6)(u + 1) = 0 Using the zero product principle gives and substituting out u gives x + 3 − 6 = 0
and x + 3 + 1 = 0 Solving x = 3 and x = −4
3.2.6
Application Problems
Example 52. A marble is dropped from the top of the Willis Tower, 442 meters above the ground.
The height of the marble is given by s(t) = −4.9t2 + 442. How long does it take the marble to reach
the ground? (Round you answer to the nearest 10th)
Solution
442
0 = −4.9t + 442 ⇐⇒ t =
=⇒ t =
4.9
2
3.3
3.3.1
2
r
442
≈ 9.5
4.9
Analyzing Graphs of Quadratic Functions
Graphing Vertex form f (x) = a(x − h)2 + k
Example 53. Find the vertex of the parabola f (x) = −5(x + 17)2 − 6.
Solution
The function is already in vertex form.
f (x) = a(x − h)2 + k
f (x) = −5(x + 17)2 − 6
Identifying a = −5,h = −17, and k = −6 gives the location of the vertex at (h, k) =⇒ (−17, −6)
Example 54. Convert to vertex form. f (x) = −3x2 + 18x − 4
Solution
College Algebra
Math 151
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Name:
Final Review
"
f (x) = −3 x2 − 6x +
−6
2
2
−
−6
2
2 #
−4 = −3[(x−3)2 −9]−4 ⇐⇒ f (x) = −3(x−3)2 +23
Example 55. Find the vertex, the axis of symmetry, the max or min and the range of f (x) =
−3x2 + 18x − 4
Solution
Convert to vertex form (see above) f (x) = −3(x − 3)2 + 23 a = −3, h = 3, and k = 23. The vertex
is at (h, k) ⇐⇒ (3, 23) a = −3 < 0 so the quadratic has a maximum. The axis of symmetry is
x = h =⇒ x = 3. The range is (−∞, 23]
3.3.2
Graphing Standard form f (x) = ax2 + bx + c
3.3.3
Vertex Formula
Example 56. Find the vertex of f (x) =
3 2
x − 6x + 5
2
Solution
h=−
3.3.4
6
−6
3
= = 2 k = f (h) = (2)2 − 6(2) + 5 = −1 ⇐⇒ (h, k) = (2, −1)
3
2
2 32
Application Problems
Example 57. A model rocket is launched with an initial velocity of 112 feet per second from a launch
pad at a height of 20 feet. The equation of motion is s(t) = −16t2 + 112t + 20. Determine when the
rocket reaches its maximum height and find the maximum height.
Solution
b
112
7
= −
= . The rocket reaches its maximum
2a
2(−16)
2
2
7
7
height after 3.5 seconds. The maximum height is s(t) = −16
+ 112
+ 20 = 216
2
2
The maximum is at the vertex. h = −
Example 58. A rubber ball is dropped from the top of a hole. Exactly 2.5 seconds later, the sound of
the rubber ball hitting the bottom is heard. How deep is the hole. Hint: The distance that a dropped
object falls in t seconds in represented by the formula s = 16t2 . The speed of sound is 1100 ft/sec.
Solution
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distance in
time two
s = 1100t2
time for the
sound to
reach the
top
distance in
time one
s = 16t21
time for the
ball to reach
the bottom
From the diagram we see that t1 + t2 = 2.5. The distance the object falls in t1 is d = 16t21 . The
distance the sound covers in t2 is d = 1100t2 . The distance is the same in both directions and
gives the equation.
16t21 = 1100t2
Using the equation t1 + t2 = 2.5 ⇐⇒ t2 = 2.5 − t1 to eliminate t2 from the equation gives
16t21 = 1100(2.5 − t1 ) ⇐⇒ 16t21 = 2750 − 1100t1
Putting the equation in standard form gives 16t21 + 1100t1 − 2750 = 0 Using the quadratic
formula with a = 16,b = 1100, and c = −2750 gives
p
√
−(1100) ± (1100)2 − 4(16)(−2750)
−1100 ± 1386000
−1100 ± 1177.285
t1 =
=
≈
2(16)
32
32
−1100 + 1177.285
= 2.41515
32
2
The distance to the bottom of the hole is s = 16(2.41515) = 93.33 ft
Rejecting the negative solution gives t1 ≈
3.4
3.4.1
Solving Rational Equations and Radical Equations
Rational Equations
Example 59. Solve
2
5
3
+
=
x2 − 9 x − 3
x+3
Solution
2
5
3
+
=
. The LCD is (x − 3)(x + 3). Clearing the fractions
(x − 3)(x + 3) x − 3
x+3
by multiplying by the LCD gives 2 + 5(x + 3) = 3(x − 3) ⇐⇒ 2 + 5x + 15 = 3x − 9 ⇐⇒ 2x =
−26 =⇒ x = −13
Factoring gives
3.4.2
Radical Equations
√
Example 60. Solve 1 − 2x − 4 = −1
Solution
Isolate the radical by adding 4 to both sides.
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√
1 − 2x = 3. Square both sides 1 − 2x = 9
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and
p solve for x.√ This gives
√ x = −4. The solution needs to be checked in the original equation.
1 − 2(−4) = 1 + 8 = 9 = 3. So x = −4 is the solution to the equation.
3.5
3.5.1
Solving Equations and Inequalities with Absolute Value
Equations with Absolute Value
Example 61. Solve 7 − |2x − 1| = 6
Solution
Isolate the absolute value. −|2x − 1| = −1 =⇒ |2x − 1| = 1. This gives the two equations
2x − 1 = −1 2x − 1 = 1. Solving each linear equation gives x = 0 and x = 1 respectively.
3.5.2
Inequalities with Absolute Value
Example 62. Solve |x + 8| < 9
Solution
This gives an “and” compound inequality −9 < x + 8 < 9 subtracting 8 from all three parts gives
−17 < x < 1 =⇒ (−17, 1)
Example 63. Solve |x + 8| > 9
Solution
This gives an “or” compound inequality x + 8 < −9 or x + 8 > 9 subtracting 8 from both sides of
both equations gives x < −17 or x > 1 (−∞, −17) ∪ (1, ∞)
College Algebra
Math 151
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Name:
4
Final Review
Chapter 4: Polynomial Functions and Rational Functions
4.1
Polynomial Functions and Modeling
Example 64. Determine the leading term, degree and classify the polynomial by type.
(a) f (x) = π 5
Solution
f (x) has leading term π 5 and it is degree 0 because it is a constant.
(b) g(x) = 1 + 4x + 3x2
Solution
g(x) has leading term 3x2 and it it degree 2, so it is a quadratic.
4.1.1
The Leading-Term Test
Example 65. Sketch the end behavior of the polynomial p(x) = x11 − 7000x2 + 1
Solution
The leading coefficient is a11 = 1 and the degree is n = 11. This gives
y
x
4.1.2
Find Zeros of Factored Polynomial Functions
Example 66. Determine whether x = 2 and x = −5 are zeros of p(x) = x3 + x2 − 17x + 15
Solution
If c is a zero of p(x), then p(c) = 0. This gives
p(2) = 23 + 22 − 17(2) + 15 = −7, So x = 2 is not a zero.
p(−5) = (−5)3 + (−5)2 − 17(−5) + 15 = 0, So x = −5 is a zero.
College Algebra
Math 151
College Algebra
Name:
Final Review
Example 67. Find the zeros of f (x) = −3(x − 3)3 (x + 2)2
Solution
Using the zero product principle gives
x−3=0
x−3=0 x−3=0 x+2=0 x+2=0
Solving each equation gives the two zeros x = 3 of multiplicity 3 and x = −2 of multiplicity 2
4.1.3
Finding Real Zeros on a Calculator
Example 68. Use a calculator to find the zero(s) of f (x) = x3 + 3x − 3
Solution
4.5
Rational Functions
4.5.1
The Domain of a Rational Function
3x + 2
Example 69. Find the domain of f (x) = 2
.
x − 5x + 6
Solution
Setting the denominator equal to zero gives
x2 − 5x + 6 = 0 ⇐⇒ (x − 3)(x − 2) = 0 =⇒ x = 3 or x = 2
This gives the domain
{x|x 6= 3 and x 6= 2} or (−∞, 2) ∪ (2, 3) ∪ (3, ∞)
4.5.2
Vertical Asymptotes
Example 70. Find the Vertical Asymptote(s) of f (x) =
College Algebra
Math 151
x−5
− 4x2
x4
College Algebra
Name:
Final Review
Solution
Setting the denominator equal to zero and solving will give the vertical asymptotes
x4 − 4x2 = 0 ⇐⇒ x2 (x2 − 4) = 0 =⇒ x2 (x − 2)(x + 2) = 0
The function has 3 vertical asymptotes x = 0, x = 2, and x = −2
4.5.3
Horizontal Asymptotes
Example 71. Find the Horizontal Asymptote of f (x) =
x5 + 7
x2 − 2x + 1
Solution
Since the degree of the numerator is 5 and the degree of the denominator is 2, the function has
no horizontal asymptote.
Example 72. Find the Horizontal Asymptote of f (x) =
3x2 − 7
5x2 + 12x − 8
Solution
Since the degree of the numerator is 2 and the degree of the denominator is 2, the horizontal
3
asymptote is the ratio of the lead coefficients. y =
5
Example 73. Find the Horizontal Asymptote of f (x) =
7x3 + 3x2 − 6
x7 − 12x5 + 3x2 − 3x + 5
Solution
Since the degree of the numerator is 3 and the degree of the denominator is 7, the horizontal
asymptote is the line y = 0.
4.5.4
Oblique (Slant) Asymptotes
Example 74. Find the oblique Asymptote of f (x) =
4x2 + 10x − 3
x+5
Solution
Using polynomial long division gives:
4x − 10
2
x+5
4x + 10x − 3
− 4x2 − 20x
− 10x − 3
10x + 50
47
The oblique asymptote is the whole part without the remainder. This gives the slant asymptote
College Algebra
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Name:
Final Review
y = 4x − 10.
4.6
4.6.1
Polynomial Inequalities and Rational Inequalities
Polynomial Inequalities
Example 75. Solve x2 − x − 5 ≥ x − 2
Solution
Setting the equation equal to zero and solving for the zeros gives
x2 − x − 5 ≥ x − 2 ⇐⇒ x2 − 2x − 3 ≥ 0 ⇐⇒ (x − 3)(x + 1) ≥ 0
Making a sign chart gives
f (x)
x−3
x+1
−
−
+
+
−
−
+
+
+
−1
x
3
f (x) is positive on (−∞, −1] ∪ [3, ∞)
4.6.2
Rational Inequalities
Example 76. Solve
x+2
≤0
x−2
Solution
Setting the equation equal to zero and solving for the zeros of the numerator and denominator
gives x = −2 and x = 2. Making a sign chart gives
f (x)
x+2
x−2
−
+
−
+
−
−
−2
+
+
+
2
x
Since x = 2 is a zero of the denominator we must exclude it. f (x) is negative on [−2, 2)
Example 77. Solve x5 + 4x3 > 4x4
Solution
Setting the equation equal to zero and then factoring gives
x5 − 4x4 + 4x3 > 0 ⇐⇒ x3 (x2 − 4x + 4) > 0 ⇐⇒ x3 (x − 2)2 > 0
The zeros of the inequality are x = 0 or x = 2 listing all of the factors and making a sign charts
College Algebra
Math 151
College Algebra
Name:
gives.
f (x) Zeros
x
0
x
0
x
0
x−2 2
x−2 2
Final Review
−
−
−
−
−
−
+
+
+
+
−
−
0
+
+
+
+
+
+
2
x
The solution set is (0, 2) ∪ (2, ∞).
College Algebra
Math 151
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5
Final Review
Chapter 5: Exponential Functions and Logarithmic Functions
5.1
5.1.1
Inverse Functions
Inverses
Example 78. Find the inverse of the relation x = y 2 − 3y + 2
Solution
To find the inverse interchange the x and the y to get y = x2 − 3x + 2
5.1.2
Inverses and One-to-One Functions
Example 79. Show y = x2 is not 1-1.
Solution
f (x)
5
4
3
2
This horizontal line intersects the curve twice
1
x
−3
−2
−1
0
1
2
3
Since a horizontal line intersects the curve more than once, it is not one to one.
5.1.3
Finding Formulas for Inverses
Example 80. Find the inverse of f (x) = 3x − 7
Solution
Replace f (x) with y
y = 3x − 7
Interchange x and y
x = 3y − 7
1
7
Solve for y
y = x+
3
3
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Math 151
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Name:
Final Review
replace y with f −1 (x)
5.1.4
f −1 (x) =
1
7
x+
3
3
Inverse Functions and Composition
Example 81. Use function composition to show that f (x) =
x+5
and f −1 (x) = 4x − 5 are inverses
4
Solution
(f ◦ f −1 )(x) =
and
(f −1 ◦ f ) (x) = 4
(4x − 5) + 5
4x
=
=x
4
4x
x+5
4
−5=x+5−5=x
So the functions are inverses.
5.2
5.2.1
Exponential Functions and Graphs
Graphing Exponential Functions
Example 82. Graph the function f (x) = 2x−2 − 5
Solution
f (x) is shifted two units to the right and down 5 units. This gives
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
5.2.2
Applications
Example 83. $100 is invested at 3% interest compounded quarterly. Find a function for the amount
in the account after t years. Use the model to find the amount of money in the account after 20 years.
College Algebra
Math 151
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Name:
Final Review
Solution
r nt
A=P 1+
n
P = 100 r = 3% = .03 n = 4 =⇒ A = 100(1.0075)4t
Using the value t = 20 gives A = 100(1.0075)80 = 181.80
5.2.3
Graphing Exponential Functions, Base e
1
Example 84. Graph the function f (x) = − ex−1 + 3
2
Solution
f (x) is shifted one unit to the left, up three units, and reflected across the x-axis. This gives
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
5.3
5.3.1
Logarithmic Functions and Graphs
Finding Certain Logarithms
Example 85. Find each of the following:
a) log 100
b) logπ π 3
1
c) log3
81
Solution
a) log 100 = 2 because 102 = 100
b) logπ π 3 = 3 because π 3 = π 3
1
1
c) log3
= −4 because 3−4 =
81
81
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Math 151
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Name:
5.3.2
Final Review
Converting Between Exponential Equations and Logarithms and Logarithmic Equations
Example 86. Convert the equation to logarithmic form 625 = 5x
Solution
log5 625 = x
Example 87. Convert the equation to logarithmic form et = 1000
Solution
ln 1000 = t
Example 88. Convert the equation to exponential form log8 x = 216
Solution
8216 = x
Example 89. Convert the equation to exponential form ln 5 = t
Solution
et = 5
5.3.3
Changing Logarithmic Bases
Example 90. Find log27 9
Solution
Using change of base on a calculator gives
log27 9 =
log 9
2
=
log 27
3
This can also be done without a calculator if the logarithm is changed to base 3
log27 9 =
5.3.4
log3 9
2
=
log3 27
3
Graphs of Logarithmic Functions
Example 91. Graph the function f (x) = − ln(x + 3)
College Algebra
Math 151
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Name:
Final Review
Solution
f (x) is shifted three units to the left, and reflected across the x-axis. This gives
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
5.4
5.4.1
Properties of Logarithmic Functions
Applying Logarithmic Properties
Example 92. Express in terms of sums and differences of logarithms log
p
x3 y 7
Solution
log
3 7
p
1
3
7
7
3
x3 y 7 = log x3 y 7 2 = log x 2 y 2 = log x 2 + log y 2 = log x + log y
2
2
Example 93. Express as a single logarithm ln(x2 − 25) − ln(x + 5) + ln(x − 5)
Solution
Working from left to right gives
x2 − 25
ln(x − 25) − ln(x + 5) + ln(x − 5) = ln
+ ln(x − 5)
x+5
2
(x − 25)(x − 5)
(x − 5)2 (x + 5)
ln
= ln
= ln(x − 5)2
x+5
x+5
2
5.4.2
Simplifying Expressions of the Type loga (ax ) and aloga (x)
Example 94. Simplify 7log7 (5t) and e3 ln x
Solution
College Algebra
Math 151
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Name:
Final Review
3
7log7 (5t) = 5t
5.5
5.5.1
e3 ln x = eln x = x3
Solving Exponential and Logarithmic Equations
Solving Exponential Equations
Example 95. Solve 5x
2
−x
= 25
Solution
Writing each side of the equation as an exponential base 5 gives
5x
2
−x
= 52 ⇐⇒ x2 − x = 2 =⇒ x2 − x − 2 = 0 ⇐⇒ (x − 2)(x + 1) = 0
This gives the two solutions x = 2 and x = −1
Example 96. Solve 21−x = 3x
Solution
ln 21−x = ln 3x ⇐⇒ (1 − x) ln 2 = x ln 3 ⇐⇒ ln 2 − x ln 2 = x ln 3
Isolating x gives
ln 2 = x ln 3 + x ln 2 ⇐⇒ ln 2 = x(ln 3 + ln 2) ⇐⇒ ln 2 = x ln 6
Solving for x gives
x=
ln 2
ln 6
Example 97. Solve 3ex − 7 = 8
Solution
3ex − 7 = 8 =⇒ 3ex = 15 =⇒ ex = 5 =⇒ x = ln 5
5.5.2
Solving Logarithmic Equations
Example 98. Solve log2 (x − 3) + log2 (x − 7) = 5
Solution
log2 (x − 3) + log2 (x − 7) = 5 ⇐⇒ log2 [(x − 3)(x − 7)] = 5
(x − 3)(x − 7) = 25 =⇒ x2 − 10x + 21 = 32
x2 − 10x + 21 = 32 =⇒ x2 − 10x − 11 = 0 =⇒ (x − 11)(x + 1) = 0
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Final Review
This gives x = 11 and x = −1. The solution x = −1 must be rejected because the logarithm of
negative number is undefined (not a real number)
5.6
5.6.1
Applications and Models: Growth and Decay; Compound Interest
Population Growth
Example 99. In a research experiment, a population of fruit flies is increasing according to the law
of exponential growth. The experiment starts with 100 flies, and after 2 days there are 300 flies. How
many flies will there be after 10 days?
Solution
Identifying P0 = 100 and P (2) = 300 gives
300 = 100e2k =⇒ 3 = e2k =⇒ ln 3 = 2k =⇒ k =
This gives P (t) = 100e(
flies
5.6.2
ln 3
2
ln 3
2
)t . So P (10) = 100e( ln23 )·10 = 24, 300 After 10 days there will be 24,300
Interest Compounded Continuously
Example 100. Aeris invests $ 100 compounded continuously. After 2 years, her account balance is
$300. How long will it take for her account balance to be $ 24,300?
Solution
Identifying P0 = 100 and P (2) = 300 gives
300 = 100e2k =⇒ 3 = e2k =⇒ ln 3 = 2k =⇒ k =
ln 3
2
This gives P (t) = 100e(ln 2 )t .
3
24300 = 100et·
ln 3
2
=⇒ 243 = et·
ln 3
2
=⇒ ln 243 = t ·
2
ln 3
=⇒ t = ln 243 ·
= 10
2
ln 3
The account balance will double after 10 years.
5.6.3
Exponential Decay
Example 101. The half-life of radioactive radium 226 Ra is 1599 years. If there are 100 mg present,
how much radioactive radium will remain after 200 years?
Solution
College Algebra
Math 151
College Algebra
Name:
Final Review
The half life can be found using the formula k =
k=
ln 2
, where T is the half-life
T
ln 2
ln 2
ln 2
=⇒ P (t) = 100e−t· 1599 =⇒ P (200) = 100e−200· 1599 ≈ 91.70
1599
So there is about 91.7 mg left after 200 years.
College Algebra
Math 151
College Algebra
Name:
9
Final Review
Chapter 9: Systems of Equations and Matrices
9.1
Systems of Equations in Two Variables
Example 102. Is (−2, 4) a solution to 5x + 2y = −2
2x − 3y = −8
Solution
We can check the ordered pair in each equation to see if it is true.
5(−2) + 2(4) = −2 True but 2(−2) − 3(4) = −16 6= −8 is False
So the ordered pair is not a solution.
9.1.1
Solving Systems of Equations Graphically
Example 103. Find the solution to the linear system
f (x)
6
4
2
x
−6
−4
−2
2
4
6
−2
−4
−6
Solution
The solution is the point of intersection of the graphs. The lines intersect at (3, 2)
9.1.2
The Substitution Method
Example 104. Solve the system
College Algebra
x+y =
5
x − 3y =
−3
Math 151
College Algebra
Name:
Final Review
Solution
Isolating x in the first equation gives x = 5 − y, substituting into the second equation gives
(5 − y) − 3y = −3 =⇒ −4y + 5 = −3 =⇒ −4y = −8 =⇒ y = 2
Substituting y back into the first equation gives x = 5 − (2) = 3. So the solution is (3, 2)
9.1.3
The Elimination Method
Example 105. Solve the system
x+y =
5
x − 3y =
−3
Solution
Multiplying the first equation by −1 and adding gives
−x
x
0
+ (−y) =
+ (−3y) =
−4y =
−5
−3
−8
Solving for y gives y = 2. Substituting y back into the first equation x + 2 = 5 =⇒ x = 3. So
the solution is (3, 2)
9.1.4
Applications
Example 106. There are a total of 50 nickels and dimes in a jar. If the value of the coins is $3.00,
how many nickels and dimes are there?
Solution
Let N be the number of nickels and D be the number of dimes, then N + D = 50. The value
is .05N + .10D = 3 =⇒ 5N + 10D = 300. Solving the first equation for N gives N = 50 − D
and substituting into the second equation gives 5(50 − D) + 10D = 300 =⇒ 250 − 5D + 10D =
300 =⇒ 5D = 50 =⇒ D = 10. So the number of nickels must be N = 40
9.2
Solving Systems of Three Equations
x +
y
Example 107. Solve 6x − 4y
5x + 2y
+
+
+
z
5z
2z
=
=
=
2
31
13
Solution
Multiplying equation 1 by −6 and adding it to equation 2 gives −10y − z = 19. Multiplying
equation 1 by −5 and adding it to equation 3 gives −3y − 3z = 3. Now multiply the new equation
College Algebra
Math 151
College Algebra
Name:
Final Review
1
3 by − . This gives
3
x +
y + z =
2
−10y − z = 19
y + z = −1
Multiply equation 3 by 10 and add to equation 2 to get 9z = 9. Then interchange the new equation
3 with equation 2 to get
x + y + z =
2
y + z = −1
9z =
9
Using back substitution gives 9z = 9 =⇒ z = 9. y + 1 = −1 =⇒ y = −2 and x − 2 + 1 = 2 =⇒
x = 3. So the solution is (3, −2, 1)
9.2.1
Mathematical Models
Example 108. Find the equation of the quadratic f (x) = ax2 + bx + c that contains the points
f (0) = 5, f (1) = 4, and f (−2) = −11
Solution
Evaluating the quadratic at the three points gives the three equations
c=5
4a − 2b + c = −11
a+b+c=4
Putting c into the last two equations gives a + b = −1 and 4a − 2b = −16. Multiplying the first
equation by 2 and adding gives 6a = −18 =⇒ a = −3 and −3 + b = −1 =⇒ b = 2. So the
equation is f (x) = −3x2 + 2x + 5
9.3
Matrices and Systems of Equations
x +
y
Example 109. Write the system as an augmented matrix. 6x − 4y
5x + 2y
+
+
+
z
5z
2z
=
=
=
2
31
13
Solution

1
 6
5
9.3.1
1
−4
2

1 2
5 31 
2 13
Gauss-Jordan Elimination
Example 110. Solve the system
3x
2x
+
+
4y
3y
=
=
22
−4
Solution
The can be solved on a TI-Graphing Calculator using the RREF command or as shown below by
hand.
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Math 151
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Name:
−R2 + R1 → R1
−2R1 + R2 → R2
−R2 + R1 → R1
Final Review
3
2
4
3
22
−4
1
2
1
3
26
−4
1
0
1
1
26
−56
1
0
0
1
82
−56
This gives (82, −56)
9.4
Matrix Operations
Most matrix operations can and should be done on a TI-Graphing Calculator. Please
refer to your notes or the tutorials in MyLabs on how to use your graphing calculator.
The Following Matrices will be used in this section


1 4 7
4
2 −5
1 4 7
−5 1 0


E=
A=
B=
C=
D= 0 1 0
1
−3 1
0 1 0
−5 4 1
1 −1 1
9.4.1
Matrix Addition and Subtraction
Example 111. Find A + D and B + C
Solution
The can be solved on a TI-Graphing Calculator.
The sum A + D is undefined. The matrices have different sizes. B and C can be added
1 4 7
−5 1 0
−4 5 7
B+C =
+
=
0 1 0
−5 4 1
−5 5 1
9.4.2
Scalar Multiplication
Example 112. Find 3D
Solution
The can be solved on a TI-Graphing Calculator or as

3 12
3D = 0 3
3 −3
College Algebra
shown below by hand.

21
0
3
Math 151
College Algebra
Name:
9.4.3
Final Review
Products of Matrices
Example 113. Multiply AE and EA
Solution
The can be solved on a TI-Graphing Calculator.
2 −5 4
3
AE =
=
−3 1
1
−11
The Product EA is undefined.
9.4.4
Matrix Equations
Example 114. Write the system
2x
−3x
+
+
3y
2y
=
=
5
as a matrix equation.
4
Solution
9.5
9.5.1
2
−3
3 x
5
=
2 y
4
2
−2
−4
3
Inverses of Matrices
The Inverse of a Matrix
Example 115. Find the inverse of A =
Solution
Use TI graphing calculator, or if you wish by hand
Create an augments matrix with the 2 × 2 Identity and A and reduce.
2 −4 1 0
−2 3 0 1
R1 + R2 → R2 and
1
R1 → R1
2
"
1
−2
0
−1
1
2
1
0
#
1
−2R2 + R1 → R1 and −1R2 → R2
"
A
College Algebra
1
0
0
1
"
−1
=
3
2
−1
−2
3
2
−1
−2
−
−
#
−1
#
−1
Math 151
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Name:
9.5.2
Final Review
Solving Systems of Equations
Example 116. Solve the system
− 4y
+ 3y
2x
−2x
=
=
6
as a matrix equation.
4
Solution
Use TI graphing calculator, or if you wish by hand
2
−2
−4
3
x
6
=
y
4
Multiplying both sides by A−1 (Found in the previous example) gives
#
# "
" 3
3
2 −4 x
6
−2
−2
−
−
=
2
2
−2
3
4
y
−1 −1
−1 −1
x
−17
=
y
−10
The solution is (−17, −10)
College Algebra
Math 151
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Name:
11
11.1
11.1.1
Final Review
Chapter 11:Sequences, Series, and Combinatorics
Sequences and Series
Sequences
Example 117. Find the 3rd and 25th term of the sequence an =
n−7
2n
Solution
a3 =
11.1.2
3−7
2
=−
2(3)
3
a25 =
25 − 7
18
9
=
=
2(25)
50
25
Finding the General Term
Example 118. Find the general term −3, 2, 7, ...
Solution
Each successive sequence member is 5 greater than the previous, this gives an = −8 + 5n
11.1.3
Sums and Series
Example 119. Find S3 for the sequence −3, 2, 7, ...
Solution
S3 = a1 + a2 + a3 = −3 + 2 + 7 = 6
11.1.4
Sigma Notation
6
X
1
Example 120. Find the sum
n
n=4
Solution
6
X
1
1 1 1
15 + 12 + 10
37
= + + =
=
n
4 5 6
60
60
n=4
Example 121. Write in sigma notation
1 1 1
1
+ + +
2 4 8 16
Solution
4
X
1
1
1
1
1
+ 2+ 3+ 4 =
2 2
2
2
2n
n=1
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Math 151
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Name:
11.2
11.2.1
Final Review
Arithmetic Sequences and Series
Arithmetic Sequences
Example 122. Find the general term of the sequence 2, 6, 10, 14, ...
Solution
The first term is a1 = 2 and the common difference is d = 6 − 2 = 4. The general term is
an = a1 + (n − 1) · d =⇒ an = 2 + (n − 1) · 4 =⇒ an = −2 + 4n
11.2.2
Sum of the First n Terms of an Arithmetic Sequence
Example 123. Find the sum of the first 100 terms of the sequence 2, 6, 10, 14, ...
Solution
a1 = 2, d = 4, n = 100, and a100 = −2 + 4(100) = 398 (See previous example) This gives
Sn =
11.2.3
100
n
(a1 + an ) =⇒ S100 =
(2 + 398) = 50 · 400 = 20000
2
2
Applications
Example 124. The sixth term of an arithmetic sequence is a6 = 54 and the thirtieth is a30 = 222.
Find the general term of the arithmetic sequence.
Solution
Method I:
There are 30 − 6 = 24 terms between the two given terms, so the common difference must
be added 24 times. This gives the equation a6 + 24d = a30 ⇐⇒ 54 + 24d = 222. Solving for d gives d = 7. Using the same technique to solve for the first terms gives the equation a1 + 5d = a6 ⇐⇒ a1 + 5(7) = 54. Solving gives a1 = 19. So the general term is
an = a1 + (n − 1) · d ⇐⇒ an = 7n + 12.
Method II: Using the formula for the general term, an = a1 + (n − 1) · d, and using the given terms
yields the system of equations.
(
54 = a1 + 5d
222 = a1 + 29d
Solving the linear system gives that a1 = 19 and d = 7. The general term is an = a1 +(n−1)·d ⇐⇒
an = 7n + 12.
11.3
11.3.1
Geometric Sequences and Series
Geometric Sequences
Example 125. Find the 5th term of a geometric sequence with a1 = 7 and r =
1
2
College Algebra
College Algebra
Math 151
Name:
Final Review
Solution
5−1
n−1
1
7
1
=⇒ a5 = 7
=
an = 7
2
2
16
11.3.2
Sum of the First n Terms of a Geometric Sequence
Example 126. Find the sum 5 −
5 5
5
+ − ... +
3 9
59049
Solution
1
a1 = 5,r = − , and n = 11. This gives
3
S11
11
5(1 − − 13 )
221435
=
=
≈ 3.75
1
59049
1 − −3
11.3.3
Infinite Geometric Series
n−1
∞
X
1
5· −
Example 127. Find (the sum)
3
n=1
Solution
a1 = 5,r = −
1
and |r| < 1 so the series converges.
3
S∞ =
11.3.4
5
5
15
= 4 =
= 3.75
4
1 − − 13
3
Applications
Example 128. Write the decimal as a fraction 0.123
Solution
123
123
123
+
+
+...
0.123 = 0.123123123... = 0.123+0.000123+0.000000123+... =
1000 1000000 1000000000
123
1
This gives that a1 =
and r =
.
1000
1000
s∞ =
College Algebra
1
123
1000
1
− 1000
=
123
1000
999
1000
=
Math 151
123
41
=
999
333
College Algebra