1 fundamentals · lecture times (all in bac 101) 1140 - 1235 1315 - 1410 1625 - 1720 1 fundamentals...
TRANSCRIPT
Lecture times (all in BAC 101)
• 1140 - 1235
• 1315 - 1410
• 1625 - 1720
1 Fundamentals (§1.1) 1
(1.1) Types of real numbers.
• Natural numbers: 1, 2, 3, 4, . . .
• Integers: . . . ,−3,−2,−1, 0, 1, 2, 3, . . .. These include the natural num-bers.
• Rational numbers: ratios of integers, e.g., 12 ,−
23 etc. These include the
integers.
• Irrational numbers: numbers that can’t be written as a ratio of inte-gers, e.g.,
√2.
(1.2) Properties of real numbers.
• Commutative properties: a+ b = b+ a and ab = ba
• Associative properties: (a+ b) + c = a+ (b+ c) and (ab)c = a(bc)
• Distributive properties: a(b+ c) = ab+ ac and (b+ c)a = ba+ ca.
(1.3) Example. Use these properties to write the following expressionswithout parentheses.
• 2(6x+ 7y) and
1Compiled November 9, 2018
1
• (a+ b)(2x+ 3y)
(1.4) Subtraction.
• Definition: a− b is defined to be a+ (−b).
• Key properties:
◦ (−1)a = −a◦ −(−a) = a
◦ (−a)b = a(−b) = −ab
◦ (−a)(−b) = ab
◦ −(a+ b) = −a− b◦ −(a− b) = b− a
(1.5) Example. Use these properties to simplify the expressions
• −2(x+ 3y − 5z) and
• (x− y)(a+ 2b− c) (use left and right distributive laws)
(1.6) Fractions and division.
• Multiplicative inverse: if a 6= 0, then 1a is the number such that a· 1a = 1.
• Definition of division: a÷ b = ab = a · 1b
• Key properties:
◦ ab ·
cd = ac
bd
◦ ab ÷
cd = a
b ·dc
◦ ac + b
c = a+bc
◦ acbc = a
b
◦ ab + c
d = ad+bcbd
◦ if ab = c
d , then ad = bc.
(1.7) Example. Use the property “acbc = a
b” to find the value of 56 + 7
15 bywriting these fraction as 30ths. (30 is the least common multiple of 6 and15.)
(1.8) Example. Use all of the properties above to simplify the expressions
• −27(28x− 14y) and
2
•(23 −
34
)(6x+ 12y)
(1.9) Ordering the real numbers:.
• a < b means b− a is positive, i.e., to the right of 0.
• a > b means b− a is negative, i.e., to the left of 0.
• a ≤ b means a < b or a = b.
• a ≥ b means a > b or a = b.
(1.10) Sets of numbers. A set is a collection of objects, usually a collectionof numbers, e.g., the set of all even integers.
(1.11) List notation. {a, b, c, . . .} is the set consisting of the numbersa, b, c, . . ..
(1.12) Example. The set of all even integers between −3 and 5 can bewritten as
{−2, 0, 2, 4}.The set of all odd natural numbers can be written as
{1, 3, 5, 7, . . .}.
(1.13) Set description notation. The notation
{x | x has property P}
denotes the set of all numbers x such that x has property P .
(1.14) Examples.
• {−2, 0, 2, 4} = {x | x is an even integer and − 3 < x < 5}
• {x | 0 ≤ x < 5} is the set of all real numbers between 0 and 5, including0 and not including 5.
(1.15) Intervals. The following are the eight different types of intervals
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• (a, b) = {x | a < x < b}
• [a, b] = {x | a ≤ x ≤ b}
• [a, b) = {x | a ≤ x < b}
• (a, b] = {x | a < x ≤ b}
• (a,∞) = {x | a < x}
• [a,∞) = {x | a ≤ x}
• (−∞, a) = {x | x < a}
• (−∞, a] = {x | x ≤ a}
(1.16) Example. (0, 2) is the set of all real numbers strictly between 0 and2. [0, 2) is the set of all real numbers between 0 and 2, including 0.
(1.17) Example. Write {x |√
4− x2 is defined} as an interval.
(1.18) Element notation.
• x ∈ A means “x is in the set A”, e.g., π ∈ [3, 4)
• x /∈ A mean “x is not in the set A”, e.g., 2 /∈ [1, 2).
(1.19) Unions. For sets A and B, the union of A and B (written A ∪ B)is the set consisting of all numbers which are either in A or in B, i.e.,
A ∪B = {x | x ∈ A or x ∈ B}.
(1.20) Example. If A = [2, 5) and B = {5}, then A ∪B = [2, 5].
(1.21) Example. Write {x |√x2 − 1 is defined} as a union of intervals.
(1.22) Intersections. For sets A and B, the intersection of A and B(written A∩B) is the set consisting of all numbers which are in both A andB, i.e.,
A ∩B = {x | x ∈ A and x ∈ B}.
(1.23) Example. If A = [2, 5) and B = [3, 7], then A ∩B = [3, 5).
(1.24) Example. Write{x∣∣∣ √x2 − 1√
4− x2is defined
}4
as a union and intersection of intervals.
(1.25) Absolute values. For a real number a,
|a| =
{a if a ≥ 0,
−a if a < 0.
(1.26) Examples. Evaluate the following absolute values
• | − 3| • |1−√
2| •∣∣3− 7
2
∣∣(1.27) Key properties.
• |a| ≥ 0
• |a| = | − a|
• |ab| = |a| |b|
•∣∣ab
∣∣ = |a||b|
• |a+ b| ≤ |a|+ |b|
2 Exponents and roots (§1.2)
(2.1) Exponents. For a real number a and an integer n > 0,
• an = a · . . . · a︸ ︷︷ ︸n times
• a−n =1
an
• a0 = 1 (if a 6= 0)
(2.2) Key properties.
• am+n = aman
• am
an= am−n
• (am)n = amn
• (ab)n = anbn
•(ab
)−n=(ba
)n•(ab
)n= an
bn
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(2.3) Remark. abn means a · (bn) not (ab)n
(2.4) Examples. Simplify the following expressions
• (2x2 + y)(y2
x − y3)
•(
(2x)3
(xy2)4
)−2
(2.5) Nth roots. y is the principal nth root of x (written y = n√x) yn = x
and, if n is even, then y ≥ 0. Note: if n is even, then x ≥ 0 also.
(2.6) Examples.
•√
4 = 2 (not −2) • 3√
8 = 2 • 3√−8 = −2
(2.7) Properties of nth roots.
• n√ab = n
√a n√b
• n
√a
b=
n√a
n√b
• m
√n√a = mn
√a
• If n is odd, n√an = a
• If n is even, n√an = |a|, e.g.,√
(−2)2 = 2
(2.8) Examples. Simplify the following expressions
•√
12−√
27
• 3√
16− 3√
54
• 3√
27x9y6
• 4
√16x2
y8
•√
16x4y2 = 4x2|y|
(2.9) Rational exponents. For a real number a and integers m and n,
am/n =(
n√a)m,
e.g.,√x = x1/2. All exponent laws above hold for rational exponents.
(2.10) Example. Simplify the following using rational exponents.
6
• 3√x
5√x2 • (x1/3 + 2x−1/3)2
3 Algebraic expressions (§1.3)
(3.1) Motivation. This section is about factoring. The main purpose offactoring is to figure out when expressions are zero. Show them an exampleinvolving height as a function of time.
(3.2) Factoring differences of squares. The key formula is
A2 −B2 = (A−B)(A+B).
(3.3) Examples. Factor these differences of squares
• x2 − 4
• 2x2 − 1
• x2 − 1x2
• 4x2 − 9y4
(3.4) Factoring perfect squares. The key formulas here are
• (A+B)2 = A2 + 2AB +B2
• (A−B)2 = A2 − 2AB +B2
(3.5) Examples. Factor these perfect squares
• 4x2 + 24x+ 9 • x2 + 6xy + 34y
2
(3.6) Factoring trinomials. The key formula is
(x+ a)(x+ b) = x2 + (a+ b)x+ ab.
(3.7) Examples. Use the formula above to factor the following expressions
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• x2 − 7x+ 12
• 2x4 + 12x2 − 14
• (x+ 1)2 − 4(x+ 1) + 3
• x2 − 6 + 7x2
(3.8) Factoring sums and differences of cubes. Observe that
• (A+B)3 = A3 + 3A2B + 3AB2 +B3
• (A−B)3 = A3 − 3A2B + 3AB2 −B3
Rearrange and factor to obtain
• A3 +B3 = (A+B)(A2 − AB +B2)
• A3 −B3 = (A−B)(A2 + AB +B2)
(3.9) Examples. Factor these sums and differences of cubes
• 8x3 + 1 • x6 − 27y3
(3.10) Important example. Group terms to factor
x3 + 2x2 − 4x− 8,
i.e., x3 + 2x2 − 4x− 8 = (x+ 2)x2 − (x+ 2)4 = etc.
(3.11) More examples. Use multiple techniques to factor these. Hint. Ineach case, factor the lowest power of x out first.
• 3x4 − 27x2
• x3/2 − 5x1/2 + 6x−1/2
• (1 + x)−2/3 + (1 + x)1/3
4 Rational expressions (§1.4)
(4.1) Examples. A rational expression is a ratio of algebraic expressions,e.g.,
8
•√x− 1
x2• 3x2 + 1
x3 − x2
(4.2) Domain of a rational expression. The domain of a rational ex-pression is the set of x for which the expression is defined, i.e., the set of xsuch that
• the numerator is defined and
• the denominator is defined and nonzero.
(4.3) Examples. Find the domains of the these rational expressions
• 1
x2 − 2x
•√x− 1
x2 − 4
• x2
x2 + 3x− 4
(4.4) Algebra of rational expressions.
• Simplifyx2 − 1
x2 − 2x− 3by canceling out common factors.
• Write1
x1/2 + 1− 2
x−1/2 − 3as a single rational expression.
• Simplifyab − 1ba + 2
. (Show them about clearing denominators as well.)
• Simplify(a+ h)−1 − a−1
h.
(4.5) Rationalizing. To rationalize the expression√a −√b, multiply by√
a+√b. Likewise, multiply
√a+√b by
√a−√b.
(4.6) Examples. Write the these expressions with no roots in the denom-inators.
9
• 1
1 +√
3•√
2
1−√
5
(4.7) Example. Rationalize the numerator of the expression
√a+ h−
√a
h.
5 Solving equations (§1.5)
(5.1) Example. I have some cows. I sell 5 cows and then sell 12 of what’s
left. Now I have 15 fewer than I started with. How many cows did I startwith? I.e., solve the equation
1
2(x− 5) = x− 15.
(5.2) Rules for solving equations.
• Add/subtract something to/from both sides.
• Multiply/divide both sides by something.
• If AB = 0, then A = 0 or B = 0.
(5.3) Linear equations. A linear equation has the form ax+ b = 0.
(5.4) Examples. Solve these equations.
• 12x+ 5 = 2
• 3x− 2 = x+ 1
(5.5) Example. Write the height of a rectangular box (with square base)in terms of its base area and surface area. (Application: fixed footprint forthe box, how tall can it be with a given amount of building material.)
(5.6) Quadratic equations. A quadratic equation has the form ax2 +bx+c = 0.
10
(5.7) Examples. Solve these quadratic equations by factoring.
• x2 − 2x− 8 = 0
• x2 + 3− 4x2 = 0
(5.8) Completing the square. Here is the general procedure for complet-ing the square.
x2 + bx+ c = 0 =⇒ x2 + bx = −c
=⇒ x2 + bx+b2
4=b2
4− c
=⇒(x+
b
2
)2
=b2
4− c
=⇒ x+b
2= ±
√b2
4− c
=⇒ x = −b2±√b2
4− c
(5.9) Example. Solve these equations for x by completing the square.
• x2 − 6x− 7 = 0
• 2x2 + 12x+ 7 = 0 (first divide by 2)
(5.10) Quadratic formula. By completing the square, it’s possible toderive the formula
ax2 + bx+ c = 0 =⇒ x =1
2a
(−b±
√b2 − 4ac
).
(5.11) The discriminant. The number
D = b2 − 4ac
in the square root above is the discriminant.
11
• D < 0 =⇒ no solutions
• D = 0 =⇒ exactly one solution (i.e., perfect square)
• D > 0 =⇒ two solutions
(5.12) Example. Solve these equations.
• x2 + 3x− 4 = 0
• x2 − x− 2 = 0
• x− 1
x= 2
• 1
x2− 3
x− 2 = 0
• x2/3 + 3x1/3 − 4 = 0
6 Graphing equations (§1.9)
(6.1) Coordinate plane. The coordinate plane is the collection of all pointsin two dimensions located by their x- and y-coordinates. (Like latitude andlongitude.) The point (0, 0) is called the origin.
(6.2) Distance formula. The distance between points (x1, y1) and (x2, y2)
is given by d =√
(x1 − x2)2 + (y1 − y2)2.
(6.3) Example. Compute the distance between (2, 3) and (1, 5). Computeeach of their distances to (0, 0).
(6.4) Graphs. The graph of an equation in x and y is the set of all points(x, y) in the coordinate plane satisfying the equation.
(6.5) Example. Graph y =1
2x by plotting points.
(6.6) Example. Graph y = x2 by plotting points.
12
(6.7) Intercepts.
• The x-intercepts of a graph are all the x such that (x, 0) is on thegraph.
• The y-intercepts of a graph are all the y such that (0, y) is on thegraph.
(6.8) Example. Find the x-intercepts of
• y = 1− x
• y = x2 − 2x
(6.9) Circles. The equation of a circle of radius R and center (k, h) is
(x− h)2 + (y − k)2 = R2.
Point out that this comes from the distance formula.
(6.10) Example. Write equations for these circles.
• Center at the origin, radius 1
• Center at (2, 3), radius√
5.
(6.11) Example. Find the x-intercepts of the circle with center (1, 1) andradius 2.
(6.12) Example. Use completing the square to find the center and radiusof the circle with equation
x2 + 2x+ y2 − y = 5.
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7 Solving inequalities (§1.8)
(7.1) Objective. Given something like 3x − 1 ≤ 2, simplify to somethinglike x ≤ 1.
(7.2) Basic rules.
• Adding/subtracting something to/from both sides won’t change aninequality.
• Multiplying both sides by a positive number won’t change an inequal-ity.
• Taking reciprocals or multiplying by a negative number flips an in-equality.
• Taking positive powers/roots of both sides won’t change an inequality.
(7.3) Example. Solve the inequality 2x+ 5 ≤ 3 for x.
(7.4) Example. Solve the inequality1
x− 1≤ 3 for x.
(7.5) Absolute value rule.
• |x| ≤ a is equivalent to −a ≤ x ≤ a. Likewise, for <.
• |x| ≥ a is equivalent to “x ≥ a or x ≤ −a. Likewise, for >.
(7.6) Examples.
• |x− 2| ≤ 3 is equivalent to 1 ≤ x ≤ 5.
• |x+ 3| ≥ 1 is equivalent to x ≥ −2 or x ≤ −4.
(7.7) Example. x2 ≤ 7 yields −√
7 ≤ x ≤√
7. (Take the square root ofboth sides and use the absolute value rule above.)
14
(7.8) Harder example. x2 − x ≤ 1 yields1
2−√
5
2≤ x ≤ 1
2+
√5
2.
(Complete the square.)
(7.9) Further examples. Solve these inequalities for x.
• 5 + x
2 + x≤ 2 (Use a sign chart.)
• x2 − x ≤ 1 (Complete the square.)
8 Equations of lines (§1.10)
(8.1) Slope. The slope of the line segment between (x1, y1) and (x2, y2) isgiven by
m =y2 − y1x2 − x1
=rise
run.
By similar triangles, any two points on a given straight line give the sameslope.
(8.2) Example. Find the slopes of the line segments between these pairsof points.
• (1, 2) and (3, 5)
• (0, 1) and (2, 0)
(8.3) Point-slope form. The line through (a, b) with slope m is given bythe equation
y − b = m(x− a).
(8.4) Example. Give an equation of the line through (−1, 4) with slope 12 .
(8.5) Example. Give an equation of the line through (1,−2) and (2, 3).
(8.6) Slope-intercept form.
15
• Simplify the equation of a line to the form y = mx+ b.
• m is the slope. b is the y-intercept.
(8.7) Example. Determine the slope and y-intercept of the line given bythe equation
7y + 5x = 2.
(8.8) Vertical and horizontal lines.
• Vertical lines have equations of the form x = a.
• Horizontal lines have equations of the form y = b.
(8.9) Example. Graph x = 2 and y = 1.
(8.10) Parallel lines. Lines are parallel if the have the same slope, or areboth vertical.
(8.11) Example. Give an equation for the line parallel to y = 3x + 5through the point (1, 9).
(8.12) Perpendicular lines. Give an equation for the line perpendicularto y = 2x+ 1 through the point (1, 3).
(8.13) Intersection points. Find the intersection point of two lines bysetting their y-values equal.
(8.14) Examples.
• Find the intersection point of y = 2x+ 3 and y = −3x+ 7.
• Find the area of the triangle enclosed by the lines
◦ y = x
◦ y = −3x+ 12
◦ y = 1
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9 Functions and graphs (§2.1, §2.2)
(9.1) Functions. A function f is a rule that assigns to each number x anumber f(x). Usually functions are described by formulas.
(9.2) Example. Suppose f(x) = x2 + 5. Evaluate f(1), f(−2) and f(3).
(9.3) Piecewise functions. Some functions are defined by cases.
(9.4) Example. You already know this one:
|x| =
{x if x ≥ 0
−x if x < 0
(9.5) Example. Let
f(x) =
{x2 + 1 if x ≥ 1
5− x if x < 1
Evaluate f(0), f(1) and f(2).
(9.6) Net change. The net change of f from a to b is f(b)− f(a).
(9.7) Example. Let f(x) = x− x2. Compute the net change of f from
• x = −1 to x = 0
• x = −1 to x = 2
(9.8) Domains. The set of permitted x-values is the domain of a function.
(9.9) Examples. Find the domain of these functions.
• f(x) =1
x2 − 1
• g(x) =√x2 − x− 1
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(9.10) Graphs. The graph of a function f is the graph of the equationy = f(x).
(9.11) Example. The graph of f(x) = 2x−1 is the line given by y = 2x−1.
(9.12) Equations. A given equation in x and y describes the graph of afunction f(x) when the equation can be solved for y uniquely in terms of x.
(9.13) Examples. Do these equations describe the graphs of functions?
• x2 + y2 = 1
• x2 + y = 1
• x2 + y3 = 1
(9.14) Vertical line test. If G is the graph of a function, each verticalline intersects G in at most one point.
(9.15) Examples. Sketch graphs of
• f(x) = x2
• f(x) = x3
• f(x) = 1/x
• f(x) = bxc (integer part of x)
• f(x) = {x} (fractional part of x) (Show them a bunch of values, e.g.,f(1.5) = 0.5, f(5.843) = 0.843, etc.)
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10 Measuring rate of change (§2.3, §2.4)
(10.1) Increasing/decreasing functions.
• A function f is increasing if x1 < x2 =⇒ f(x1) < f(x2)
• A function f is decreasing if x1 < x2 =⇒ f(x1) > f(x2).
(10.2) Examples. Graph these functions to see where they are increasingand decreasing.
• f(x) = x− x2
• f(x) = x3 − 3x2 + 1 (Local extrema at x = 0, 2)
(10.3) Local extrema.
• The value f(c) is a local maximum of f if f(x) ≤ f(c) for all x near c.
• The value f(c) is a local minimum of f if f(x) ≥ f(c) for all x near c.
(10.4) Example. Draw them a picture and point out the local extrema.
(10.5) Average rate of change. The average rate of change of f from ato b is
f(b)− f(a)
b− a.
(10.6) Remark. Should be called “average slope”. Illustrate this with apicture.
(10.7) Examples. If you drive 150 miles in 5 hours, your average velocityis 30 mph. This ignores any change in velocity.
(10.8) Example. Compute the net change and average rate of change of
• f(x) =1
2x2 +
3
2from x = 1 to x = 3
19
• g(x) =x
x− 1from x = 2 to x = 5
(10.9) Examples.
• Compute the average rate of change of h(x) = mx+ c
◦ from x = 2 to x = 5 ◦ from x = a to x = b
• Compute the average rate of change of f(x) = x2
◦ from x = 1 to x = 3/2
◦ from x = 1 to x = 5/4
◦ from x = 1 to x = b
◦ from x = a to x = b
• Compute the average rate of change of g(x) =√x
◦ from x = 4 to x = 5
◦ from x = 4 to x = 9/2
◦ from x = 4 to x = b
◦ from x = a to x = b
• Compute the average rate of change of f(x) = 1/(x2 + 1)
◦ from x = 1 to x = 3
◦ from x = 2 to x = 3
◦ from x = b to x = 1
◦ from x = a to x = b
11 Combining functions (§2.6, §2.7)
(11.1) Sums. The sum/difference of functions f and g is the function(f ± g)(x) = f(x)± g(x).
(11.2) Example. Write
h(x) = x2 − 1
x+ 1
as a difference of simpler functions.
20
(11.3) Products. The product/quotient of f and g is the function (fg)(x) =f(x)g(x) (or (f/g)(x) = f(x)/g(x).
(11.4) Example. Write h(x) = x2√x+ 1 as a product of simpler functions.
(11.5) Composition. If f and g are functions, the composite of f and g isf ◦ g given by (f ◦ g)(x) = f(g(x)).
(11.6) Remark. In general, f ◦ g 6= g ◦ f .
(11.7) Examples. Evaluate and simplify f ◦ g and g ◦ f for these pairs offunctions.
• f(x) = x+ 1 and g(x) = x2
• f(x) =√x+ 1 and g(x) = 2x2
• f(x) =1
x+ x2 and g(x) = x+ 1
(11.8) Examples. Write these functions in the form f ◦ g.
• h(x) =√
2x2 + 1
• h(x) =1
x2 − 5
(11.9) Example. Let f(x) =√x, g(x) = x2 + 1 and h(x) =
1
x. Evaluate
these combinations of functions.
• (f ◦ g)(x) + (h ◦ f)(x)
• (h ◦ g)(x)
f(x)
• f(x)(g ◦ h)(x).
(11.10) Rules for graphing function combos.
• f(x) 7→ f(x) + 3 shifts the graph of f up 3.
21
• f(x) 7→ 2f(x) stretches the graph of f vertically by a factor of 2.
• f(x) 7→ f(3x) shrinks the graph horizontally by a factor of 3.
• f(x) 7→ f(x− 5) shifts the graph of f right by 5.
• f(x) 7→ f(x+ 5) shifts the graph of f left by 5.
• f(x) 7→ −f(x) flips the graph of f vertically.
• f(x) 7→ f(−x) flips the graph of f horizontally.
(11.11) Examples. Graph these function combinations.
• Let f(x) = 2x+ 1 and simplify and graph
◦ f(x− 3)
◦ f(x+ 1)
◦ f(2x)
◦ f(x) + 3
• Let f(x) =1
xand graph
◦ f(x− 3) ◦ f(x)− 1 ◦ f(2− x)
(11.12) Example. Let f(x) = −x2 + 8x − 13. Complete the square andtreat f as a modification of g(x) = x2 in order to graph it.
22
12 Inverse functions (§2.8)
(12.1) Example. Let f(x) = 2x+ 1.
x f(x)0 11 3? 5
• Based on the pattern, the “?” is 2.
• In general, is there a formula to get x back from f(x)?
• Solve for x in the equation y = 2x+ 1
• This yields g(y) = 12y −
12
• Notice that g(1) = 0, g(3) = 1 and g(5) = 2, etc.
(12.2) Inverses. A function g is the inverse of f if g(f(x)) = x andf(g(x)) = x. The notation for the inverse of f is “f−1”.
(12.3) Example. (a non-invertible function) Let f(x) = x2. If g is aninverse for f , then, since f(2) = 4
g(f(−2)) = g(4) = g(f(2)) = 2 6= −2.
(12.4) One-to-one functions. For f to have an inverse, it must be one-to-one, i.e., x1 6= x2 =⇒ f(x1) 6= f(x2).
(12.5) Examples. Solve for the inverses of these functions.
• f(x) = 3x− 5
• f(x) =1
x+ 1
• f(x) =x
2x+ 1
• f(x) =x+ 2
x+ 3
• f(x) =√
2x+ 3
• f(x) =
√x
1 +√x
23
(12.6) Example. (partial inverse) Find an inverse for f(x) = 2x − x2 onthese intervals:
• [1,∞) ...it works out to f−1(y) = 1 +√
1− y
• (−∞, 1] ... it works out to f−1(y) = 1−√
1− y
13 Polynomials (§3.1, §3.2)
(13.1) Quadratic polynomials. A quadratic polynomial has the form
f(x) = ax2 + bx+ c
(13.2) Maximum and minimum values. The maximum or minimum
value of f(x) = ax2 + bx+ c occurs wen x = − b
2a(i.e., at the vertex).
• a > 0 =⇒ it’s a minimum.
• a < 0 =⇒ it’s a maximum.
(13.3) Example. Find the maximum value of f(x) = 2x− x2.
(13.4) Example. You have 1000 feet of fence material, what’s the largestyard you can enclose?
(13.5) Example. You have 1000 feet of fence material, what’s the largestyard you can enclose if one edge is along the side of a river?
(13.6) Example. You sell computers at price x. At x = $800, you sell 10computers per week. With every $100 increase in the price, you sell 1 lesscomputer per week. What price will maximize your revenue?
revenue = (# sold)× (price)
= (10− x− 800
100)x
etc.
24
(13.7) Higher degree polynomials. Here is some terminology by exam-ple:
−3x5 + 7x4 − 2x3 + 9x2 + x+ 7.
Point out the
• degree
• leading coefficient/term
• coefficients
• constant term
(13.8) End behavior. There are four cases:
• Even degree with positive leading coefficient.
• Even degree with negative leading coefficient.
• Odd degree with positive leading coefficient.
• Odd degree with negative leading coefficient.
Draw a picture for each case.
(13.9) Zeroes. A number c is a zero of a function f if f(c) = 0. If n is thedegree of f , then f has at most n zeroes.
(13.10) Multiplicity. If f is a polynomial function, then c is a zero off exactly when (x − c) is a factor of f . The multiplicity of a zero c is thelargest m such that (x− c)m divides f(x).
(13.11) Example. Let f(x) = x3 + 3x2 − 9x− 27. Notice that 3 is a zeroof multiplicity 1 and −3 is a zero of multiplicity 2. Use this and knowledgeof the end behavior to sketch a graph of this polynomial.
(13.12) Intermediate value theorem. If f is a polynomial function andf(x1) and f(x2) have opposite signs, then f has a zero between x1 and x2.
(13.13) Example. Use the IVT to approximate√
2.
25
14 Graphing rational functions (§3.2, §3.6)
(14.1) Range of a function. The range of a function f is the set of allpossible y-values, i.e., {f(x) | x ∈ domain(f)}.
(14.2) Example. The range of f(x) = 3− x2 in (−∞, 3].
(14.3) Graphing near a zero. Suppose that c is a zero of f with multi-plicity m. Show them what the graph of f looks like near a zero when
• m = 1
• m > 1 and m is odd
• m > 1 and m is even
(14.4) Local maxima and minima.
• If f has degree n, then f has a most n− 1 local maxima and minima.
• A polynomial function f has at least one local maximum or minimumvalue between zeroes.
(14.5) Example. Graph f(x) = 4x− x3.
(14.6) Example. Graph f(x) = −(x− 1)2(x− 3)(x− 5).
(14.7) Rational functions. A rational function is a ratio of polynomials,e.g.,
f(x) =x2 + 1
3x3 + x+ 2.
(14.8) Vertical asymptotes.
• Show them examples by picture.
• The line x = a is a vertical asymptote of f if |f(x)| gets larger andlarger as x gets closer to a.
• Vertical asymptotes come from zeroes of the denominator.
26
(14.9) Example. Find the vertical asymptotes of f(x) =1
x2 − 1, sketch
its graph and determine the range of f .
(14.10) Example. Find the vertical asymptotes of f(x) =x
x2 + x− 6,
sketch its graph and determine the range of f .
(14.11) Horizontal asymptotes. If the numerator and denominator of arational function f have the same degree, then f(x) approaches the ratio ofthe leading coefficients as |x| get larger. If r is this ratio, the line y = r is ahorizontal asymptote of f .
(14.12) Example. Sketch the graph of f(x) =(2x− 1)(x− 1)
(x− 2)(x+ 3). Use a sign
chart. Determine the range of f .
(14.13) Other end behavior. Suppose f is a rational function where thenumerator has degree m and the denominator has degree n.
• m > n =⇒ |f(x)| gets larger and larger as |x| gets larger and larger.
• m < n =⇒ |f(x)| gets closer to zero as |x| gets larger and larger.
(14.14) Example. Sketch the graph of f(x) =2x3
2− x2and determine its
range.
27
15 Exponential functions (§4.1, §4.2)
(15.1) Exponential functions. An exponential function is a function ofthe form f(x) = ax, where a is a fixed positive number.
(15.2) Example. Sketch the graph of f(x) = 2x by plotting some samplepoints. Notice that f(x+ 1) = 2f(x). So when x increases (or decreases) by1, f(x) increases (or shrinks) by a factor of 2. Also note the end behaviorof f . This is typical of exponential functions.
(15.3) Example. Sketch the graph of g(x) = (1/2)x in two ways.
• Plot some points.
• Let f(x) = 2x and notice that g(x) = f(−x). So the graph of g is thegraph of f reflected across the y-axis.
(15.4) Example. Sketch the graph of p(x) = 1−1/3x using transformationsand the fact that 1/3x = 3(−x).
(15.5) Compound interest. Suppose you put $1000 in a bank accountwhich has an annual interest rate of 3% compounded quarterly. This meansthat every 3 months your balance increases by 3
4%. Thus, after one year,you have
1000
(1 +
0.03
4
)·(
1 +0.03
4
)·(
1 +0.03
4
)·(
1 +0.03
4
)
= 1000
(1 +
0.03
4
)4
≈ $1030.34
If it’s compounded monthly, you have
1000 ·(
1 +0.03
12
)12
≈ $1030.41
after one year. The more frequently you compound, the more money youget.
28
(15.6) Compound interest formula. In general, if you start with Pdollars and have an annual interest rate of r, compounded n times per year,the amount of money you have after t years is
money = P(
1 +r
n
)nt(15.7) Example. Suppose you make 100% interest, compounded annually.After one year, you have 200% of the money you started with. What if youcompound it more frequently?
Two times per year...
(1 +
1
2
)2
≈ 225%
Three times per year...
(1 +
1
3
)3
≈ 237%
Four times per year...
(1 +
1
4
)4
≈ 244%
(15.8) The natural exponential. As n gets larger,
(1 +
1
n
)n
≈ e ≈ 2.71828...
This is corresponds to compounding interest more and more frequently.
(15.9) Example. Population is often modeled by functions like
p(t) =1500e2t
e2t + 3
Based on this, explain why the maximum population is 1500.
29
16 Logarithms (§4.3, §4.4, §4.5)
(16.1) Logarithm. The function loga(x) is defined as the inverse functionof ax, i.e.,
loga(ax) = aloga(x) = x.
Write loge(x) = ln(x).
(16.2) Laws of logarithms. These are derived from the correspondinglaws for exponents
• loga(xy) = loga(x) + loga(y)
• loga(x/y) = loga(x)− loga(y)
• loga(xp) = p · loga(x)
• loga(a) = 1
• loga(1) = 0
• logb(x) =loga(x)
loga(b)...change of base formula
(16.3) Example.
• log10(10, 000) = log10(104) = 4
• log2(32) = log2(25) = 5
• ln(73x) = 3x ln(7)
• log10(4 · x3) = log10(4) + 3 log1 0(x)
• ln(2/e3x) = ln(2)− 3x
• ln(6x4/(7y−2)) = etc.
• log2(2x/43x) = etc.
30
(16.4) Graph of the logarithm. Sketch the graph of f(x) = loga(x).Explain what the domain, range and end behavior are.
(16.5) Examples. Use transformations to sketch the graphs of the follow-ing functions.
• f(x) = 1 + ln(x− 2)
• f(x) = 2− ln(3− x)
• f(x) = ln(|x− 2|) =
{ln(x− 2) if x > 2
ln(2− x) if x < 2
(16.6) Solving equations. Factor these exponential polynomials and uselogarithms to solve for x.
• e2x − 1 = 0
• e2x − 3ex + 2 = 0
• e3x − 3e2x − 4ex + 12 = 0
31
17 Trigonometric functions (§5.1, §5.2)
(17.1) The unit circle. The unit circle is the circle of radius 1, centeredat (0, 0). Its circumference is 2π.
(17.2) Measuring angles. The starting point on the unit circle is (1, 0).Given a real number α > 0, measure distance α CCW along the perimeterof the circle to arrive and the associated terminal point. If α < 0, measuredistance CW around the circle.
(17.3) The four quadrants.
• x, y > 0 =⇒ quadrant I
• x < 0, y > 0 =⇒ quadrant II
• x, y < 0 =⇒ quadrant III
• x > 0, y < 0 =⇒ quadrant IV
Draw a picture as well.
(17.4) Example. Indicate the terminal points associated with
• α = π
• α = 2π
• α = −π/3
• α = 2π/3
• α = −11π/6
• α = 13π/2
(17.5) Definitions of sine and cosine.
• cos(α) is the x-coordinate of the terminal point associated with α.
• sin(α) is the y-coordinate of the terminal point associated with α.
Indicate when sine and cosine are positive vs. negative in the four quadrants.
32
(17.6) Table of “cardinal” values. Here are the most important valuesof sine and cosine for 0 ≤ α ≤ π/2.
α 0 π/6 π/4 π/3 π/2
sin(α) 0 1/2√
2/2√
3/2 1
cos(α) 1√
3/2√
2/2 1/2 0
(17.7) Examples. Use the table above to compute these values of sine andcosine.
• sin(2π/3)
• cos(5π/4)
• sin(7π/6)
• cos(5π/3)
(17.8) Other trig functions.
• tan(α) =sin(α)
cos(α)
• sec(α) =1
cos(α)
• cot(α) =cos(α)
sin(α)
• csc(α) =1
sin(α)
(17.9) Negative angle formulas.
• sin(−α) = − sin(α)
• cos(−α) = cos(α)
• tan(−α) = − tan(α)
• cot(−α) = − cot(α)
• sec(−α) = sec(α)
• csc(−α) = − csc(α)
(17.10) Notation. sinn(α) = (sin(α))n
(17.11) Pythagorean identities.
• sin2(α) + cos2(α) = 1
• tan2(α) + 1 = sec2(α)
33
(17.12) Examples.
• Assume sin(α) = 3/5 and 0 ≤ α ≤ π/2. What are the values of cos(α)and tan(α)?
• Assume cos(α) = −4/7 and π ≤ α ≤ 3π/2. What are the values ofsin(α) and tan(α)?
34
18 Graphs of trig functions (§5.3, §5.4)
(18.1) Sine and cosine. Sketch the graphs of sine and cosine for the class.Have a unit circle on the side so they can see when the functions are positivevs. negative. Also point out that
−1 ≤ sin(α) ≤ 1 and − 1 ≤ cos(α) ≤ 1
for all α.
(18.2) Periodicity. You can see from the graphs that
cos(α + 2nπ) = cos(α) and sin(α + 2nπ) = sin(α)
for each integer n.
(18.3) Example. Sketch the graph of f(x) = sin(π/2 + x). It’s the sameas cosine.
(18.4) Sample value chart. Use the following style of value chart whengraphing composites.
x sin(x) sin2(x)0 0 0
π/4√
2/2 1/2π/2 1 1
etc.
(18.5) Examples. Make a table of values and sketch the graphs of
• f(x) = sin2(x)
• f(x) = 1− cos2(x)
• f(x) = cos(2x)
Point out the relationship between the first and the last.
(18.6) Examples. Sketch the graphs of
35
• f(x) = x sin(x)
• f(x) = x2 sin(x)
• f(x) = e−x sin(x)
(18.7) Secant and tangent. Sketch the graphs of secant and tangent on[0, 2π], noting the vertical asymptotes. Use a sign chart (as with rationalfunctions) to help with the graphing.
(18.8) Examples. Sketch the graphs of
• f(x) = tan(π − x)
• f(x) = sec(2x)
• f(x) = cot(x)
• f(x) = csc(x)
• f(x) = tan2(x)
• f(x) = sec2(x)
36
19 Inverse trig functions (§5.5)
(19.1) Inverse functions. If f is a one-to-one function,
f−1(y) = the x such that f(x) = y.
E.g., f(x) = 2x =⇒ f−1(x) = log2(x). To see this in action, recall thatlog2(8) = 3 because 3 is the x such that 2x = 8.
(19.2) Problem. Trigonometric functions aren’t one-to-one so how do Italk about inverse trigonometric functions. Instead use “partial” inverses.
(19.3) Inverse sine.
• Sketch the graph of y = sin(x).
• As an example, sin−1(√
2/2) should be “the” α such that
sin(x) =√
2/2,
but this could be π/4, 3π/4, etc.
• Define sin−1(x) to be the α such that
◦ sin(α) = x and
◦ −π/2 ≤ α ≤ π/2.
• Thus, for instance, sin−1(√
2/2) = π/4, not 3π/4.
(19.4) Inverse cosine.
• Sketch the graph of y = cos(x).
• Define cos−1(x) to be the α such that
◦ cos(α) = x and
◦ 0 ≤ α ≤ π.
37
(19.5) Examples.
• cos−1(cos(5π/4)) = 3π/4
• cos(sin−1(−1/2)) =√
3/2
• sin(cos−1(−1/7)) = 4√
3/7
Solution: let α = cos−1(−1/7) and note that α is in quadrant II. Solvefor sine using the first Pythagorean identity (not the triangle).
(19.6) Inverse tangent.
• Sketch the graph of y = tan(x).
• For instance, notice that tan−1(0) could be 0, π, etc.
• Define tan−1(x) to be the α such that
◦ tan(α) = x and
◦ −π/2 < α < π/2.
• Sketch a graph of y = tan−1(x) and note the HAs: y = π/2 on theright and y = −π/2 on the left. Justify the HAs based on the graphof tangent.
(19.7) Examples.
• tan−1(−1) = −π/4
• tan−1(−√
3) = −π/3
• tan−1(1/√
3) = π/6
• cos(tan−1(8)) = 1/√
65
Solution: Let α = tan−1(8). Use the second Pythagorean identity toget sec(α). Then get cos(α). Don’t use the triangle yet.
• sin(tan−1(8)) = 8/√
65. (Use the previous problem.)
38
20 Working with triangles (§6.2, §6.3)
(20.1) Degree vs. arc length measure. With degree measure of angles,the full circle is divided into 360 degrees, instead of 2π. For instance,
• 360◦ ∼ 2π
• 180◦ ∼ π
• 270◦ ∼ 3π/2
• 90◦ ∼ π/2
• 60◦ ∼ π/3
• 120◦ ∼ 2π/3
(20.2) Example. (The 30◦ − 60◦ − 90◦ triangle)
• Draw the 30-60-90 triangle inside the unit circle and an arbitrary 30-60-90 with opposite, adjacent and hypotenuse labeled
• By similar triangles,
◦ sin(30◦) = 1/2 = opp./hyp.
◦ cos(30◦) =√
3/2 = adj./hyp.
◦ tan(30◦) = (1/2)/(√
3/2) = opp./adj.
(20.3) Trig functions via triangles.
• Draw a right triangle with base angle α. Label base angle, oppositeside, adjacent side and hypotenuse.
• In general, (for 0 ≤ α < 90◦),
◦ sin(α) =opposite
hypotenuse
◦ cos(α) =adjacent
hypotenuse
◦ tan(α) =opposite
adjacent
39
(20.4) Example. Evaluate these expression using triangles.
• tan(sin−1(1/3))
• sin(tan−1(3/2))
• tan(sin−1(−2/5))
• sec(tan−1(−7/4))
(20.5) Figuring out the 30-60-90 triangle.
• Draw the 30-60-90 triangle with hypotenuse 1. Double it to get anequilateral with side lengths 1.
• Use Pythagorean theorem to get the height.
• Solve for sine and cosine.
(20.6) Figuring out the 45-45-90 triangle.
• Draw a 45-45-90 triangle with hypotenuse 1.
• Use the fact that base equals height to solve for the side lengths.
• Solve for sine and cosine.
(20.7) Example. Suppose you are standing 50’ from the base of a reallytall tree. The top of the tree is 60◦ above the ground from your point ofview. How tall is the tree? (50
√3 ≈ 87′)
(20.8) Example. Suppose a triangle has two edges with lengths 3 and 7,and the angle between these edges is 70◦. What is the area of the triangle?
(20.9) Example. Suppose a triangle has one side of length 5, oppositeangle of 40◦ and intervening height equal to 4. What is the area of thetriangle?
40
21 Some trig identities (§6.4, §7.1, §7.3)
(21.1) Strategy. When simplifying expression with lots of trig functions,one approach is to write everything in terms of sine and cosine and see whatcancels.
(21.2) Examples. Simplify these expressions.
• sin(x)
tan(x)
• tan(x)
sec(x)
• cos2(x) sec(x) tan2(x)
• tan2(x)
sin3(x)
(21.3) Examples. Simplify these expressions.
• cos(x)
sec(x)+
sin(x)
csc(x)
• 2 + tan2(x)
sec2(x)− 1 ...use the fact that 1 + tan2(x) = sec2(x)
• 1− cos(x)
sin(x)+
sin(x)
1− cos(x)...the answer is 2 csc(x)
(21.4) Example. Verify that tan2(x)− sin2(x) = tan2(x) sin2(x).
(21.5) Examples. Use triangles to simplify these expressions.
• cos(sin−1(x))
• tan(cos−1(x))
• sin2(tan−1(x))
• sin(cos−1(x))
(21.6) Double-angle formulas.
• sin(2x) = 2 sin(x) cos(x)
• cos(2x) = cos2(x)− sin2(x)
• tan(2x) =2 sin(x) cos(x)
cos2(x)− sin2(x)· 1/ cos2(x)
1/ cos2(x)=
2 tan(x)
1− tan2(x)
41
(21.7) Example. Simplify these expressions.
• cos(2 sin−1(x)) ...the answer is 1− 2x2
• tan(2 sin−1(x)) ...the answer is2x(1− x2)1/2
1− 2x2
(21.8) Half-angle formulas.
• sin2(x) = 12(1− cos(2x))
• cos2(x) = 12(1 + cos(2x))
• tan2(x) =1− cos(2x)
1 + cos(2x)
(21.9) Examples. Find exact values.
• cos(π/12)
• sin(π/8)
• tan(π/12)
(21.10) Example. Simplify sin2(12 tan−1(x))
42
22 More trig equations (§7.2, §7.4)
(22.1) Angle addition and subtraction formulas.
• sin(α + β) = sin(α) cos(β) + cos(α) sin(β)
• sin(α− β) = sin(α) cos(β)− cos(α) sin(β)
• cos(α + β) = cos(α) cos(β)− sin(α) sin(β)
• cos(α− β) = cos(α) cos(β) + sin(α) sin(β)
(22.2) Proof. (Subtraction formula for cosine) Compare these pairs ofpoints
• (cos(α− β), sin(α− β)) and (1, 0)
• (cos(α), sin(α)) and (cos(β), sin(β))
Each pair covers the same angular distance. Hence, they are the samedistance apart. Use distance formula, square both sides and simplify.
(22.3) Example. From last time,
cos( π
12
)=
√2 +√
3
2.
Use the cosine subtraction formula to compute cos(π/12) using the fact thatπ/12 = π/3 − π/4. Check equality between the two answers by squaringeach.
(22.4) Example. Compute sin(π/12).
(22.5) Example. Evaluate these expressions.
• cos(11π/12) = cos(π − π/12)
• sin(7π/12) = sin(π/2 + π/12)
• cos(5π/12) = cos(π/3 + π/12)
• sin(17π/12) = sin(3π/2− π/12)
43
(22.6) Example. Simplify these expressions.
• sin(cos−1(x2) + tan−1(x/5))
• cos(tan−1(x/3)− sin−1(x/6))
• cos(32 sin−1(x)) = cos(sin−1(x) + 12 sin−1(x))
(22.7) Solving trig equations. An equation like
sin(x) = 1
is a matter of finding those x whose sine is 1. More complicated equationsmay require factoring.
(22.8) Example. Find all x with 0 ≤ x ≤ 2π which satisfy the followingequations.
• 2 sin2(x)− sin(x)− 1 = 0
• 2 sin2(x)− 1 = 0
• 4 cos2(x)− 4 cos(x) + 1 = 0
• 2 cos2(x)− 1 = 0
• 2 cos2(x) sin(x)− sin(x) = 0
• 2 sin3(x)− sin(x) = 0
• 4 cos2(x)− 3 = 0
• 4 sin2(x) cos(x)− 3 cos(x) = 0
44
23 Systems of equations (§10.1, §10,2)
(23.1) Example. Find the intersection point of the lines x − y = 2 and3x+ 2y = 1. Use elimination and point out the method.
(23.2) Example. I have 11 quarters and dimes. The total value is $1.85.How many of each do I have?
(23.3) General approach. Given the system of equations
ax + by = rcx + dy = s
there are two steps:
• Subtract ca × (equation 1) from equation 2.
• Solve for x by subbing y into the first equation.
(23.4) Example. Solve this system of equations.
2x + 3y = 45x − 7y = 3
(23.5) Example. Find a two digit number m such that
• the digits of m sum to 7 and
• reversing the digits of m increases the number by 27.
(23.6) Example. Two H2SO4 solutions: one is 10% concentration and theother is 70%. I want 3 liters of 60% H2SO4. How much of each solutionshould I use?
(23.7) Example. (Three variables) Use two-step elimination to solve thissystem.
x + 2y − 3z = 13x + y + z = 02x − 2y + z = 1
45
(23.8) Example. (A system with many solutions)
x + y + z = 1x − y + 2z = 32x + 3z = 4
(23.9) Example. (A system with no solutions)
x + y + z = 1x − y + 3z = 32x + 4z = 5
(23.10) Example. (Diet) Given the following nutritional information
food fat (g per 100) protein (g) carbs (g)P(eanut butter) 48 24 0
B(read) 3 8 20C(hicken) 0 30 0
set up a system of equations which tells you how much of each ingredient toget a 4000 kCal diet with 30/30/40 kCals from fat/pro/carbs.
46
24 Partial fractions (§10.7)
(24.1) Example. Notice that
1
x2 − 1=
1/2
x− 1− 1/2
x+ 1
Given1
x2 − 1, how would you figure this out? Assume it can be done, i.e.,
1
x2 − 1=
A
x− 1+
B
x+ 1
and solve for A and B. This is called a partial fractions decomposition.
(24.2) Example. Find the partial fractions decomposition of
2x− 1
x2 − 1
(24.3) General idea.
• Start withp(x)
q(x)– where p and q are polynomials.
• Factor q(x) into (x− λ1), . . . , (x− λk).
• Writep(x)
q(x)=
A1
x− λ1+ . . .+
Ak
x− λkI’ll explain later what happens if there are irreducible quadratics.
• Solve for A1, . . . , Ak.
47
(24.4) Example. Find the partial fractions decomposition these rationalfunctions.
• 1
x2 − 3x+ 2
• 2x− 1
x2 − 3x+ 2
• x+ 5
x2 + 6x+ 8
• 2x2 + x− 3
x3 + 2x2 − 9x− 18
(24.5) Example. (Irreducible quadratic factors) Find the partial fractionsdecomposition these rational functions.
• x2 + 2x− 1
x3 + x=Ax+B
x2 + 1+C
x
• 2x2 + 3
(x2 + 2)(x− 1)
(24.6) Example. (Repeated factors) Find the partial fraction decomposi-tion of these rational functions.
• x+ 1
(x− 1)2(x+ 2)=
A
x− 1+
B
(x− 1)2+
C
x+ 2
• x2 + x+ 1
(x+ 3)2(x− 1)
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