Intro
f
consists of a set of inputs, a set of outputs, and a rule for assigning each input to exactly one output.A
and B
, a set with elements that are ordered pairs (x, y)
, where x
is an element of A
and y
is an element of B
, is a relation from A
to B
.f
with domain D
, we often use x
to denote the input and y
to denote the output associated with x
.x
as the independent variable and y
as the dependent variable, because it depends on x
.y = f(x)
, and we read this equation as “y equals f of x.”(x, y)
in the coordinate plane where y = f(x)
. The graph of a function is the set of all these points.x
for which f(x)
is a real number.{ x | 1 > x < 5 }
or(1, 5)
f
, the graph of f
is the set of points (x, f(x))
, where x
is in the domain of f
and f(x)
is in the range.x
where f(x) = 0
are called the zeros of a function.f
intersects the x-axis.(0, f(0))
.f
, every vertical line that may be drawn intersects the graph of f
no more than once. If any vertical line intersects a set of points more than once, the set of points does not represent a function.f
is increasing on the interval I
if for all x₁, x₂ ∈ I
,
f(x₁) ≤ f(x₂)
when x₁ < x₂
.f
is strictly increasing on the interval I
if for all x₁, x₂ ∈ I
,
f(x₁) < f(x₂)
when x₁ < x₂
.f
is decreasing on the interval I
if for all x₁, x₂ ∈ I
,
f(x₁) ≥ f(x₂)
when x₁ < x₂
.f
is strictly decreasing on the interval I
if for all x₁, x₂ ∈ I
,
f(x₁) > f(x₂)
when x₁ < x₂
.f(x) = ax + b
, where a
and b
are constants.y
for each unit change in x
.y
versus the change in x
. To do so, we choose any two points (x₁, y₁)
and (x₂, y₂)
on the line and calculate Slope(m)=ΔxΔy=x2−x1y2−y1.(x₁, y₁)
and (x₂, y₂)
on the line.f(x) = ax + b
:f
at x = 0
, we see that (0, b)
is a point on this line.x = 1
, we see that (1, a + b)
is also a point on this line.f(x) = mx + b
tells us the slope, m
, and the y-intercept, (0, b)
. This form of expressing a linear function is called the slope-intercept form.(x₁, y₁)
with slope m
. The equation y − y₁ = m (x - x₁)
is the point-slope form for that linear function.ax + by = c
, where a
and b
are both not zero. This form is more general because it allows for a vertical line
, x = k
.f(x) = mx + b
is a polynomial of degree 1 if m ≠ 0
and degree 0 if m = 0
.f(x) = 0
is called the zero function and its degree is undefined.f(x) = ax2 + bx + c
, where a ≠ 0
.f(x) = axᵇ
, where a
and b
are any real numbers.b
is a positive integer.b
is even, then f(x) = axᵇ
is an even function because f(−x) = a(−x)ᵇ = axᵇ
if b
is even.b
is odd, then f(x) = axᵇ
is an odd function because f(−x) = a(−x)ᵇ = −axᵇ
if b
is odd.f(x)
is what happens to the values of f(x)
as x → ∞
and as x → −∞
.f(x)
either:c
and we say “f(x)
approaches c
as x
goes to infinity,” and we write f(x) → c
as x → ∞
. Also, the line y = c
is a horizontal asymptote for the function. E.g, f(x) = 2 + 1/x
.f(x) → ∞
as x → ∞
. E.g, f(x) = 3x²
.f(x) = ax² + bx + c
.a > 0
, the values f(x) → ∞
as x → ±∞
.a < 0
, the values f(x) → −∞
as x → ±∞
.a > 0
; the parabola opens downward if a < 0
.f(x) = ax³ + bx2 + cx + d
.a > 0
, then f(x) → ∞
as x → ∞
and f(x) → −∞
as x → −∞
.a < 0
, then f(x) → −∞
as x → ∞
and f(x) → ∞
as x → −∞
.f
intersects the x-axis, we need to solve the equation f(x) = 0
for x
:f(x) = mx + b
, the x-intercept is given by (−b/m, 0)
.ax² + bx + c = 0
:x = (−b ± √(b2 − 4ac)) ÷ 2a
. The discriminant part of the formula is b² − 4ac
.b² − 4ac > 0
, there are two real numbers that satisfy the quadratic equation.b² − 4ac = 0
, there is only real number one solution.b² − 4ac < 0
, no real numbers satisfy the quadratic equation.f(x) = p(x) / q(x)
, where p(x)
and q(x)
are polynomials.f(x) = x^1/n
, where n
is a positive integer greater than one.n
is even, the domain of f(x) = x^1/n
is [0, ∞)
.n
is odd, the domain of f(x) = x^1/n
is [-∞, ∞)
and the function is an odd function.x < a
and x > a
, we need to pay special attention to what happens at x = a
when we graph the function. Sometimes the graph needs to include an open or closed circle to indicate the value of the function at x = a
. An open circle is used to denote that a graph doesn’t define the function output at x = a
. A closed circle is used for the opposite effect.y = f(x)
, we can apply multiple function transformations to it: y = cf(a(x + b)) + d
. When function transformations are combined, you can follow the order below to get the transformed function’s graph from the base function’s graph:y = f(x)
.y = f(x + b)
by a factor of |a|
. If a < 0
, reflect the graph about the y-axis.y = f(a(x + b))
by a factor of |c|
. If c < 0
, reflect the graph about the x-axis.y = cf(a(x + b))
.An enumeration of the different function transformations and their related effects on the graph of a function:
| Transformation of f
; Where c > 0
| Effect on the graph of f
|
|--------------------------------------|--------------------------------------------------------------|
| f(x) + c
| Vertical shift up c
units |
| f(x) − c
| Vertical shift down c
units |
| f(x + c)
| Shift left by c
units |
| f(x − c)
| Shift right by c
units |
| cf(x) | Vertical stretch if c > 1
;
Vertical compression if 0 < c < 1
|
| f(cx) | Horizontal stretch if 0 < c < 1
;
Horizontal compression if c > 1
|
| −f(x) | Reflection about the x-axis |
| f(−x) | Reflection about the y-axis |
Almost any repetitive or cyclical motion can be modeled by some combination of trigonometric functions.
Radians are a more natural measurement of angles compared to degrees because they are related directly to the unit circle, a circle with radius 1.
The radian measure of an angle θ
is the arc length s
of the associated arc on the unit circle. I.e. the angle corresponding to the arc of length 1 has radian measure 1.
Since an angle of 360°
corresponds to the circumference of a circle, or an arc of length 2π
(on the unit circle), we conclude that an angle with a degree measure of 360°
has a radian measure of 2π
.
Use the fact that 180° is equivalent to π radians as a conversion factor:
1 = π rad / 180° = 180° / π rad
Let P = (x, y)
be a point on the unit circle centered at the origin O
. Let θ
be an angle with an initial side along the positive x-axis and a terminal side given by the line segment OP
. The trigonometric functions are then defined as:
sinθ = y
cscθ = 1/y
cosθ = x
secθ = 1/x
tanθ = y/x
cotθ = x/y
If x = 0
, secθ
and tanθ
are undefined. If y = 0
, then cotθ
and cscθ
are undefined.
For a point P = (x, y)
on a circle of radius r
with a corresponding angle θ
, the coordinates x
and y
satisfy:
cosθ = x / r
x = r / cosθ
sinθ = y / r
y =r / sinθ
LThe ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions evaluated at either of the acute angles of the triangle. Let θ
be one of the acute angles. Let A
be the length of the adjacent leg, O
be the length of the opposite leg, and H
be the length of the hypotenuse. By inscribing the triangle into a circle of radius H
, we see that A
, H
, and O
satisfy the following relationships with θ:
sinθ = O / H
cosθ = A / H
tanθ = O / A
cscθ = H / O
secθ = H / A
cotθ = A / O
A trigonometric identity is an equation involving trigonometric functions that is true for all angles θ
for which the functions are defined. We can use the identities to help us solve or simplify equations.
The main trigonometric identities are:
tanθ = sinθ / cosθ
cotθ = cosθ / sinθ
cscθ = 1 / sinθ
secθ = 1 / cosθ
The trigonometric functions are periodic functions.
The period of a function f
is defined to be the smallest positive value p
such that f(x + p) = f(x)
for all values x
in the domain of f
.
The sine, cosine, secant, and cosecant functions have a period of 2π
. Since, the angle θ
and θ +2π
correspond to the same point on the unit circle.
Since the tangent and cotangent functions repeat on an interval of length π
, their period is π
.
[Figure1.34]
Transformations can be applied to trigonometric functions. E.g. f(x) = A sin(B(x − α)) + C
where:
α
causes a horizontal or phase shift.B
changes the period. This transformed sine function will have a period 2π / |B|
.A
results in a vertical stretch by a factor of |A|
. We say |A|
is the amplitude of f.C
causes a vertical shift.
Figure1.35The graph of y = cosx
is the graph of y = sinx
shifted to the left π / 2
units: cosx = sin(x + π/2)
.
Similarly, we can view the graph of y = sinx
as the graph of y = cosx
shifted right π/2
units: sinx = cos(x − π/2)
.
f
to elements in the range of f
. The inverse function maps each element from the range of f
back to its corresponding element from the domain of f
.f
with domain D
and range R
, its inverse function (if it exists) is the function f⁻¹
with domain R
and range D
such that f⁻¹(y) = x
if f(x) = y
. In other words, for a function f
and its inverse f⁻¹
:f⁻¹(f(x)) = x
for all x
in D
, andf(f⁻¹(y)) = y
for all y
in R
.f⁻¹
is read as f inverse. The -1
is not used as not used as an exponent.f
becomes the domain of f⁻¹
and the domain of f
becomes the range of f⁻¹
.f
is a one-to-one function if f(x1) ≠ f(x2)
when x1 ≠ x2
.f
is one-to-one if and only if every horizontal line intersects the graph of f
no more than once.f
:y = f(x)
for x
.x
and y
and write y = f⁻¹(x)
. This is because by convention, x
represent the independent variable and y
represent the dependent variable. Representing the inverse function in this way is also helpful when graphing a function f
and its inverse f⁻¹
on the same axes.f
and the graph of its inverse:f
and a point (a, b)
on the graph.b = f(a)
, then f⁻¹(b) = a
.f⁻¹
, the point (b, a)
is on the graph.f⁻¹
is a reflection of the graph of f
about the line y = x
.f
we can choose a subset of the domain of f
such that the function is one-to-one. This subset is called a restricted domain. By restricting the domain of f
, we can define a new function g
such that the domain of g
is the restricted domain of f
and g(x) = f(x)
for all x
in the domain of g
. Then we can define an inverse function for g
on that domain.sin⁻¹
or arcsin
, and the inverse cosine function, denoted cos⁻¹
or arccos
, are defined on the domain D = {x| − 1 ≤ x ≤ 1}
as follows:sin⁻¹(x) = y
if and only if sin(y) = x
and −π/2 ≤ y ≤ π/2
;cos⁻¹(x) = y
if and only if cos(y) = x
and 0 ≤ y ≤ π
.tan⁻¹
or arctan
, and inverse cotangent function, denoted cot⁻¹
or arccot
, are defined on the domain D = {x| − ∞ < x < ∞}
as follows:tan⁻¹(x) = y
if and only if tan(y) = x
and − π 2 < y< π 2
;cot⁻¹(x) = y
if and only if cot(y) = x
and 0 < y < π
.csc⁻¹
or arccsc
, and inverse secant function, denoted sec⁻¹
or arcsec
, are defined on the domain D = {x |x| ≥ 1}
as follows:csc⁻¹(x) = y
if and only if csc(y) = x
and − π 2 ≤ y≤ π 2
, y ≠ 0
;sec⁻¹(x) = y
if and only if sec(y) = x
and 0 ≤ y ≤ π, y ≠ π/2
.[−π/2, π/2]
. By doing so, we define the inverse sine function on the domain [−1, 1]
such that for any x in the interval [−1, 1]
, the inverse sine function tells us which angle θ in the interval [−π/2, π/2]
satisfies sinθ = x
.f(x) = bˣ
, where b > 0
, b ≠ 1
, is an exponential function with base b
and exponent x
.(−∞, ∞)
and the range is (0, ∞)
. b > 1
, f(x) = bˣ
is increasing on (−∞, ∞)
and bˣ → ∞
as x → ∞
, whereas bˣ → 0
as x → −∞
.0 < b < 1
, f(x) = bˣ
is decreasing on (−∞, ∞)
and bˣ → 0
as x → ∞
, whereas bˣ → 0
as x → −∞
.f(x) = xᵇ
for some constant b
is not an exponential function but a power function.x
is a positive integer, then bˣ = b · b ⋯ b
(with x
factors of b
).x
is a negative integer, then bˣ = 1/bˣ
.x
is zero, then bˣ = 1
.x
is a rational number (x = c/d
; where c
and d
are integers), then bˣ = ᵈ√bᶜ
.bˣ · bʸ = bˣ⁺ʸ
bˣ / bʸ = bˣ⁻ʸ
(bˣ)ʸ = bˣʸ
(ab)ˣ = aˣbˣ
aˣ / bˣ = (a/b)ˣ
f(x) = eˣ
is called the natural exponential function.e ≈ 2.718282
.f(x) = eˣ
is the only exponential function with tangent line at x=0
that has aslope of 1
.b > 0
, b ≠ 1
, the logarithmic function with base b
, denoted logb, has domain (0, ∞)
and range (−∞, ∞)
, and satisfies logb(x) = y
if and only if bʸ = x
.