Chapter 3 Functions
First: Describe the set of inputs, and the set of outputs.
1. The cost of a tank of gasoline depends on the number of gallons purchased.
2. The volume of a cube depends on the length of its edges.
Function – contains a set of inputs, called the domain, and a set of outputs called the
range. Differs from a relation because each input determines one and only one
output (x coordinate is never repeated – but y may be)
a) Function Notation – f(x) = some function
b) Used when want to evaluate a function
Ex. f(x) = 3x² + 2 for f(3)
3(3)² + 2 = 29
Ex. g(x) = √2x² + 7
a) g(1)
b) g(-3)
c) g(0)
Difference Quotient - if f is a function, then the quantity f(x+h) – f(x)
h
**Rule: replace x with x+h then subtract the original function and divide by h
Ex. f(x)= 3x²-2x use the difference quotient
3(x+h)² - 2(x+h) –(3x² - 2x) = 3x²+6xh+3h² - 2x-2h -3x²+2x =6xh+3h²-2h
h
h
h
= h(6x+3h-2) = 6x+3h-2
h
Functions defined by equations:-- equations in 2 variables can be used to define
function,
But not every equation in 2 variables represents a function.
Steps: 1) check to see if can be solved for x – and that exactly 1 value of y is
produced.
Ex. 2y³-32x + 11 = 0
Ex. 2x + y²-3 = 0
Domains – will consist of every real number input for which the function rule produces a
real number output
Except (scenarios where domains won’t consist of all real #’s)
1) division by 0
2) square root of a - # or nth root when n is even
Ex. find the domain for the given function: (write in interval notation)
a) f(x) = 3
x+4
b) f(x) = √1 – 3x
Piecewise-Defined Function – one whose rule includes several formulas
Ex. f(x) 3x² - 5 if x ≥4
2x + 15 if x<3
Find each of the following:
F(0)—since 0 is less than 3—2nd rule applies 2(0)+ 15 = 15
F(3.5)
F(4)
Greatest Integer Function – is a piecewise defined function in which for any # x, round
down to the nearest integer less than or equal to x.
 domain is always all real numbers, and range is the set of integers
 notation is f(x) = [x]
Ex. let f(x) = [x] Evaluate the following:
a) f(-209.99) - [-209.99] = -210
b) f(-51) – [-51] = -51
c) f(-5/4)
d) f(√10)
-----------------------------------------------------------------------------------------------------------3.2 Graphs of functions
x coordinate = input
Do Ex. 1 p. 150
y coordinate = output
Vertical Line Test – a graph in a coordinate plane represents a function if and only if no
vertical line intersects the graph more than once.
** remember – this is because a function can’t repeat its x-coordinate
Analyzing Graphs:
a) Increasing and Decreasing Functions
1) Increasing = if its graph always rises as you move from left to right in
interval
2) Decreasing = if its graph always falls as you move from left to right in
interval
3) Constant on Interval – if its graph is a horizontal line over the interval
Ex. on what interval is the function f(x) = │x+3│- │x│increasing,decreasing,con.
b) Local Maxima and Minima – the peaks and valleys of a graph
**make sure that your viewing window is adjusted enough to see peaks/valleys
Ex. graph f(x) = x³-2.1x²+x+3 and find all local maxima and minima
c) Concavity and Inflection points – used to describe the way a curve bends
1. Concave up – for any 2 points in an interval- if the segment that connects
them is above the curve
2. Concave down –if the segment is below the curve
3. Inflection Point – a point where the curve changes concavity
Ex. Graph f(x) = x³ + 3x²-x find the maxima, minima, intervals where increasing and
decreasing, inflection points, where concave up and down
Graphs of Piecewise Functions – often discontinuous with jumps or holes, graph each
piece separately
ex. graph f(x) = 3
for -2 ≤ x ≤ 2
x²-4x for x>2
a) Absolute Value Function – is a piecewise function b/c definition is:
│x│= -x if x<0
x if x>0
b) Greatest Integer Function – is a piecewise function b/c each value represents
an interval.
Ex. graph f(x) = [x]-2
(start by picking intervals, -3≤x<-2, etc)
Parametric Graphing – the x-coordinate and y-coordinate of each point on a graph are
each given as a function of the third variable, t, called a parameter.
The functions that give the rules for the coordinates = parametric equation
a) f(t) = (x,y)
x = x(t) y = y(t)
b) to graph by hand make a table for t, x and y – then plug t into x equation, then t
into y equation. Outputs = coordinate point
Ex. graph curve given by: x = t-2
y= t² + 3t
t
-3
-2
x=t-2
y=t²+3t
(x,y)
0
1
2
** to graph in calculator – you must change your mode to parametric
3.3 Quadratic Functions – any polynomial of degree 2, and whose shape is a parabola
Axis of Symmetry- a line through the vertex that makes the parabola symmetric
Quadratics can be written in 2 forms:
1. Transformation Form
f(x) = a(x-h)²+ k
a) most useful for finding the vertex, b/c has points (h,k)
b) y-intercept = in forms of ah²+k – putting a 0 in for x and getting y
c) x- intercepts – put a o in for y and solve for x – are h±√-k/a
ex. For function f(x)= -4(x-2)² + 5 – find the vertex, intercepts, graph
2. Polynomial Form
f(x) = ax²+bx+c
a) most useful for find y-intercept – put 0 in for x y will always = c
b) x-intercepts- solution of quad. Eq.- so find by factoring or quad.form
c) vertex – use formula h = –b
2a ** once get h, plug into equation to get k
Ex. f(x) = x²+4x+5 – find the vertex, x and y intercepts, graph
3. X-intercept form
f(x) = a(x-s)(x-t)
a) most useful for finding the x-intercepts = set both = 0 and solve (s, and t)
b) y intercept – put 0 in for x and solve for y = ast
c) vertex- is exactly half way between the x intercepts so x= s + t
2
**to get y coordinate, just plug x into equation
Ex. f(x) = -2(x-3)(x+1) find vertex, and intercepts, then graph
Changing from 1 form to another:
1. to change form either transformation or x intercept form to polynomial form –
distribute and combine like terms.
2. To change either polynomial or transformation form to x intercept form, factor
out leading coefficient then factor/use quad.form. to find x intercepts
Applications – the vertex of a parabola is a max or min of a function so answers
application questions.
Ex. Find the dimensions of a rectangular parking lot that can be enclosed with 2400 feet
of fence and that has the largest possible area.
--let x denote the length and y the width so:
Perimeter = 2x + 2y = 2400 solve for y 2y=2400-2x then y = 1200 - x
Area = xy
plug in: A=x(1200-x)= 1200x - x²
Now use formula for vertex: -b
-1200
=600
2a
2(-1)
Now plug in to get y: y = 1200 – 600 = 600 so vertex = (600,600)
3.4 Graphs and transformations:
Parent Functions – a function with a certain shape that has the simplest algebraic rule for
that shape.
a)
b)
c)
d)
e)
f)
g)
h)
i)
#1. Vertical Shifts
Constant f(x) =1
Identity f(x)=x
Absolute Value f(x) = │x│
Greatest Integer f(x) = [x]
Quadratic f(x) = x²
Cubic f(x) = x³
Reciprocal f(x) = 1/x
Square root f(x) = √x
Cube root f(x) = 3√x
g(x)= f(x) + c is graph shifted up c units
g(x) = f(x) – c is graph shifted down c units
Ex. Graph f(x) = │x│ then graph f(x) = │x│+ 3 then f(x) = │x│-2
#2. Horizontal Shifts g(x) = f(x+c) is shifted c units to left
g(x) = f(x-c) is shifted c units to right
Ex. Graph: f(x) = x² then graph f(x)= (x-3)² then graph f(x) = (x+4)²
#3 Reflection g(x) = f(-x) is the graph reflected across y-axis
g(x) = -f(x) is the graph reflected across x-axis
Ex. graph: g(x) = x²
and
g(x) = -x²
#4 Vertical Stretches and Compressions: if (x,y) is a point on the graph of f(x) then
(x,cy) is a point on the graph of g(x)=c∙f(x)
a) if c>1 the graph of f is stretched vertically, away from x axis by factor of c
b) if c <1 the graph of f is compressed vertically toward the x axis by factor of c
** main thing to remember – changing y coordinate**
Ex. graph f(x) = 2│x│ and h(x) = ½│x│
#5. Horizontal Stretches and Compressions – if (x,y) is a point on the graph of f(x) then
(1/cx,y) is a point on the graph g(x) = f(c∙x)
a) if c>1 the graph is compressed horizontally toward y axis by factor of 1/c
b) if c<1 the graph is stretched horizontally away from y axis by factor of 1/c
**main thing to remember – changing x coordinate**
Ex. Graph f(x) = 33√3
and
g(x) = 3√3
Combining Transformation: for a function of form g(x) = c∙f(a(x-b)) + d, first graph f(x)
1. if a<0 reflect the graph across the y axis
2. stretch or compress the graph horizontally by a factor of │1/a│
3. Shift the graph horizontally by b units, right if b>0 left if b<0
4. If c<0 reflect the graph across the x axis
5. Stretch or compress the graph vertically by a factor of │c│
6. Shift the graph vertically by d units up if d>0, down if d<0
Ex. graph g(x) = -(3x -12)² + 2
a)first: rewrite factored out: -(3(x-4))²+2 so know the parent function
b) second: compress horizontally by factor of 1/3
c) third: shift 4 units to the right
d) fourth : reflect across x axis
e) fifth: shift upward 2 units
3.4 Symmetry
3 kinds of symmetry:
#1. Y-AXIS symmetry- occurs if the part of the graph on the right side is the mirror
image of the part of the graph on the left side of the y-axis.
**is symmetric with respect to the y-axis if whenever (x,y) is on the graph,
then (-x,y) is on the graph – algebraically replacing x by –x produces equivalent eq.
a) So y-coordinates are always the same
b)The x-coordinates are opposites of each other
c) The y-axis will be a perpendicular bisector between pts. P and Q
Ex. verify that y = -2x4 + 3x2 is symmetric with respect to the y-axis
Steps: plug in –x wherever there is an x
Y = -2(-x)4 + 3(-x)² = -2x4 + 3x² which is equivalent to original
Therefore, symmetric about y axis.
#2. X- AXIS Symmetry – a graph is symmetric with respect to the x-axis if whenever
(x,y) is on the graph, then (x,-y) is also on it. – algebraically
this means that replacing y by –y produces an equivalent
equation.
a) x-coordinates are always the same
b) their y-coordinates are opposite of each other
c) The x-axis is the perpendicular bisector between pts. P and Q
Ex. Verify that y²= x + 2 is symmetric with respect to the x-axis
Steps: plug in –y wherever there is a y in the equation
(-y)² = x + 2 = y²=x + 2 which is equivalent to original equation
So is symmetric about x-axis
#3. Origin Symmetry – a graph is symmetric with respect to the origin if whenever (x,y)
is on the graph, then (-x,-y) is also on it—algebraically this
means replacing x by –x and y by –y produces an equivalent eq
a) if a line through the origin and any point P on the graph also intersects the
graph at point Q such that the origin is the midpoint of PQ
Ex. verify that y = -x³ + 5x is symmetric with respect to the origin
2
Steps: replace x by –x and y by –y, then simplify
-y = -(-x)³ + 5(-x)
2
-y = x³ - 5x
= y = -x³ + 5x
-1 2(-1) (-1)
2
Even and Odd Functions:
A. Even Functions – a function f is even if f(-x) = f(x) for every value x in the domain
of f. The graph of an even function is symmetric with respect to
the y-axis.
Ex. f(x) = x4 + x² + 3
f(-x) = (-x)4 + (-x)² = x4 + x² = f(x) is the same
f(x) = f(-x)
B. Odd Functions – a function f is odd if f(-x) = -f(x) for every value x in the domain of f
The graph of an odd function is symmetric with respect to the
origin.
Ex. f(x) = 4x
-f(x) = -(4x)
so -f(x) = f(-x) so odd function
3.5 Operations on Functions: ways in which 2 or more functions can be combined to
make new functions.
1. Sums and Differences of functions
f+g(x) = f(x) + g(x) **is basically combining of like terms
f-g(x) = f(x) – g(x)
**Domain and Range of sum and difference is all real numbers that are in
the domain of each function.
Ex. f(x) = √1-x and g(x) = √x² - 1 find the following and domain
Find (f+g)(x) = √1-x + √x²-1
**can’t combine any like terms
(f-g)(x) = √1-x - √x²-1
**can’t combine any like terms
Domain: for a) domain is all x such that
2. Product and Quotient Functions are defined by the following:
(fg)(x) = f(x) ∙g(x)
and f(x) = f(x)
g
g(x)
Domain for these consist of all real #’s in both the domain of f and g
Ex. f(x) = √2x and g(x) =√x²-4
find (fg)(x) and f/g(x)
(fg)(x) = √2x(√x²-4) = √2x³ - 8x
(f/g)(x) = √2x
∙ √x²-4 = √2x³-8x
√x²-4
√x²-4
x²-4
Domain: of f: x≥0
Of g: x≤-2 or x ≥2
Of fg x≥2
Of f/g x > 2
3. Products with a constant function: distribute to each item in function
a) domain will be the same as domain of f
Composition of Functions – using the output of 1 function as the input of the other
** If f and g are functions then the composite function of f and g is (g◦f)(x) = g(f(x))
 the expression g◦f is read “g circle of f” or “f followed by g”
 order of function – applied right to left (inner to outer)
Ex. f(x) = x² + 2x -1 and g(x) = 1 find the following:
x-3
a) (g◦f)(-2)
b) (f◦g)(4)
c) (g◦f)(x)
d) (f◦g)(x)
Domains of Composite Functions:
Let f and g be function. The domain of g◦f is the set of all real numbers x such
that:
a) x is in the domain of f
b) f(x) is in the domain of g
Ex. if f(x) = √x-2 and g(x) = x² - 1 find
a) g◦f = (√x-2)² - 1 = x-2-1 = x-3
b) f◦g = (√x²-1-2) = (√x²-3)
c) domain of g◦f = domain of f is x≥2, the domain of g is all real # - the domain
of f(x) is in g, so the domain is x≥2
d) domain of f◦g = domain of g is all real #’s, the values of g(x) that are in the
domain of f are :√x²-3≥0 = √x² ≥3 take square root of each side and domain
is: x≤-√3 or x≥√3
Expressing Functions as composites
 taking a composite and breaking into its functions
h(x) = √x³-5 write h as a composition of function in two different ways
g(x) = √x
f(x) = x³ - 5
3.6 Inverse Functions
Inverse Function – the result of exchanging the input and output values of a function
a) (x,y) is a point on the graph of a function ----- (y,x) is a point on its inverse
b) the reflection of the graph of the original function across the line y=x
Do. Ex. 1 p. 205 --- just need to reverse coordinates
Finding Inverses from an Equation:
EX. find g(x), the inverse of f(x) =2x+1
Step 1: write the functions in terms of x and y
y = 2x + 1
Step 2: Switch the x and y
x= 2y + 1
Step 3: Now solve for y (so presented in function notation)
x= 2y + 1
-1
-1
x – 1 = 2y
2
2
so: g(x) = x-1
2
Determing whether an Inverse is a function
1)** every input of the inverse corresponds to exactly one output (x’s can’t be repeated)
= one-to-one function
** if the original function is one-to-one, then its inverse is also a function
2) Horizontal Line Test – a function is one-to-one, if and only if no horizontal line
Intersects the graph of f more than once
Ex. Graph each function below and determine whether the function is one-to-one,
a) f(x) = 3x5 -2x4 + 3x³ + x -2
b) g(x) = -.5x³ + x + 1
Let f be a function. The following statements are equivalent:
 The inverse of f is a function
 f is one-to-one
 The graph of f passes the horizontal line test
The inverse function, if it exists, is written as f-1 where if:
y = f(x) then x = f-1(y)
Restricting the Domain: done when a function is not one-to-one. It is possible to
produce an inverse function by considering only a part of the
function that is one-to-one
Ex. Find an interval on which the function f(x) = x²-3 is one to one, and find f-1 on that
Interval:
Step 1: graph f(x) =x² -3 ** looking at graph see 1:1 on interval [0,∞)
Step 2: switch the x and y x = y² - 3
Step 3: solve
√x+3 = √y² --- y = ±√x+3
So f-1(x) = √x+3 b/c want the nonnegative coordinates
Composition of a Function and its Inverse
A one-to-one function f and its inverse function f-1 have these properties:
(f-1 ◦f)(x) = x
for every x in the domain of f
-1
(f◦ f )(x) = x
for every x in the domain of f-1
Any two functions having both properties are one-to-one and are inverses of each other.
Ex. Let f(x) = 2x
3-x
and
g(x) = 3x
x+2
Use composition to verify that f and g are
inverses
3.7 Rates of change
When dropping a rock from a high place, the distance the rock travels is given by the
function: d(t) = 16t².
Fill in the table:
Time (t)
0
1
2
3
3.5
4
4.5
5
Distance
d(t)
Find the distance the rock falls from t=1 to t = 3
In general: the distance traveled from time t = a to time t = b is d(b) – d(a) feet
The length of time interval from t = a to t= b is b – a seconds
Therefore:
average speed = distance traveled
Time interval
d(b) – d(a)
b–a
(measured in ft/sec)
Ex. Using above formula:
find the average speed of the falling rock from t = 2 to t = 5
Overall, average speed is the most familiar example, but in general:
Average Rate of Change = change in f(x)
Change in x
=
f(b) – f(a)
b–a
Ex. a cone-shaped tank is being filled with water. The approximate volume of
Water in the tank in cubic meters is V(x) = x³/4, where x is the height of the
Water in the tank. Find the average rate of change of the volume of water as
The height increases from 1 to 3 meters.
Geometric Interpretation:
Secant line – a straight line that joins 2 points
Secant Line and Average Rate of Change =
Average rate of change of f from x=a to x =b
f(b) – f(a)
b-a
will give you the slope of the secant line joining (a, f(a)) and (b, f(b)) on graph f
=
The difference quotient – used when computing the rate of change over an interval from
a to a+h for some small quantity h. The difference quotient allows us to define a formula
to determine the average rate for all possible values of h
Using the difference quotient to find the average rate of change: f(x+h) – f(x)
h
Ex. The distance traveled by a dropped object is given by the function
D(x) = 4.9x² with distance d(x) measured in meters and time in seconds. Find
A formula for the average speed of a falling object from time x to time x+h.
Use the formula to find the average speed from 2.8 to 3 seconds.
Ex. find the difference quotient of v(x) = x³ to find the average rate of change of
The volume of a cube whose side has a length of x as x changes from:
a. 4 to 4.1
b. 4 to 4.01
c. 4 to 4.001
Steps: first use the difference formula for x³.