Chapter 1
Chapter 1
Chapter 1
1 Functions
• This section deals with the topic of
functions, one of the most important topics
in all of mathematics. Let’s discuss the
idea of the Cartesian coordinate system
first.
II I
(3,1)
x
(-1,-1)
III IV
1
Graphing an equation
• To graph an equation in x and y, we need to find ordered
pairs that solve the equation and plot the ordered pairs on
a grid.
2
Function
• The previous graph is the graph of a function. The idea of
a function is this: a relationship between two sets D and R
• such that for each element of the first set, D, there
corresponds one and only one element of the second set, R.
Function definition
• You can visualize a function by the following diagram which shows
a correspondence between two sets, D, the domain of the function
and R, the range of the function. The domain gives the diameter of pizzas
and the range gives the cost of the pizza. range
domain
10
9.00
12 16 10.00
12.00
3
Function Notation
• The following notation is used to describe functions The
variable y will now be called
Function evaluation
• Consider our function
Some Examples
• 1.
f (a) = 3(a) − 2
f (6 + h) = 3(6 + h) − 2 = 18 + 3h − 2
= 16 + 3h
4
Domain of a Function
• Consider
f ( x) = 3 x − 2
f (0) = ?
f (0) = 3(0) − 2 = −2
Domain of a function
• Answer:
Domain of a function
• Therefore, the domain of our function is the set of real
numbers that are greater than or equal to 2
3
• Answer:
{ x x ≥ 8} , [8, ∞)
5
More examples
• Find the domain of 1
f ( x) =
3x − 5
Mathematical modeling
• The price-demand function for a company is given by
•
p ( x) = 1000 − 5 x, 0 ≤ x ≤ 100
• where P(x) represents the price of the item and x
represents the number of items. Determine the revenue
function and find the revenue generated if 50 items are
sold.
Solution
• Revenue = price x quantity so
• R(x)= p(x)*x = (1000 − 5 x)i x
• When 50 items are sold, x = 50 so we will evaluate the
revenue function at x = 50
6
1.2 Elementary Functions; Graphs and
Transformations
• In this presentation, you will be given an equation of a
function and asked to draw its graph. You should be able
to state how the graph is related to a “standard” function.
It is not important that you plot a great many points for
each graph. It IS important that you recognize the general
shape of the graph. You can verify your answers using a
graphing calculator, but only after you have attempted to
construct the graph by hand.
Problem 1
• Construct the graph of
Solution
12
10
8
6 Series1
4
2
0
-4 -2 0 2 4
1
Problem 2
• Now, sketch the related graph given by the
equation below and explain, in words, how it is
related to the first function you graphed.
Solution: Problem 2
• The graph has the same shape as the original
function. The difference is that the original graph
has been translated two units to the right on the
x-axis. Conclusion: The graph of the
function f(x-2) is the graph of f(x) shifted
horizontally two units to the right on the x-axis.
• Notice that replacing x by x-2 shifts the graph
horizontally to the right and not the left.
Problem 3
• Now, graph the following “standard”
function: Complete the table:
-3
-2
-1
0
1
2
3
2
Solution to problem 3
30
20
10
0 Ser ies1
-4 -2 0 2 4
- 10
- 20
- 30
Problem 4
• Now, graph the following related function:
Solution to problem 4
30
20
10
0 Ser ies1
-4 -2 0 2 4
-10
- 20
- 30
3
Problem 4 solution
• The graph of
Problem 5
• Graph:
Solution to problem 5
• The domain is all non-negative real
numbers. Here is the graph:
6
5
4
3 Series1
2
1
0
0 5 10 15 20 25 30
4
Problem 6
• Graph:
Problem 6 solution
(Notice that the graph lies entirely within the fourth
quadrant)
0
0 5 10 15 20 25 30
-1
-2
-3 Series1
-4
-5
-6
Graph of –f(x)
• The graph of the function –f(x) is a reflection of
the graph of f(x) across the x-axis. That is, if the
graphs of f(x) and –f(x) are folded along the x-
axis, the two graphs would coincide.
5
Cube root function
• Sketch the graph of the cube root function. Complete the
table of ordered pairs: x y
-27
-8
-1
0
1
8
27
6
Same graph as graph of cube root function. Shifted
horizontally to the left one unit.
a ( x) = x
7
Graph of absolute value
function
Notice the symmetry of the graph.
a ( x + 1) − 2 = x + 1 − 2
Shift absolute value graph to the left one unit and down two
units on the vertical axis.
8
Linear functions and Straight Lines
Linear Functions
• The equation f(x) = mx+b m and b are real
numbers is the equation of a linear
function. The domain is the set of all real
numbers. The graph of a linear function is
a straight line. Some examples of graphs
will follow in the next few slides.
1
f ( x) = x+8
4
1
f(x)= -2x+3
More examples
f ( x ) = −2
Graphing
• Graph 3
f ( x) = x+2
4
• using a table of
values for x and y
x y
-4
0
4
8
2
Solution:
solution
3
Special cases
• 1. The graph of x=k is the graph of a
vertical line k units from the y-axis.
Some examples:
• 1. Graph x=-7
• 2. Graph y = 3
solutions
X=7
Y=3
4
Slope of a line
y2 − y1
• Slope of a line: m=
x2 − x1
D
( x1 , y1 )
• = rise
run
Rise
D ( x2 , y2 )
run
Slope-intercept form
• The equation y = mx + b
5
Point-slope form
• The point- slope form of
the equation of a line is
as follows: y − y1 = m( x − x1 )
Examples
• Find the equation of the line through the
points (-5, 7) and (4, 16) :
• Solution:
( −5, 7 ) (4,16)
16 − 7 9
m= = =1
4 − (−5) 9
y − 16 = 1( x − 4) → y = x − 4 + 16 = x + 12
Applications
• Office equipment was purchased for $20,000 and will
have a scrap value of $2,000 after 10 years. If its value
is depreciated linearly , find the linear equation that
relates value (V) in dollars to time (t) in years:
6
Quadratic functions
• If a, b, c are real numbers with a not equal
to zero, then the function
f ( x) = ax 2 + bx + c
• is a quadratic function and its graph is a
parabola.
f ( x ) = a ( x − h) 2 + k
Consider f ( x ) = −3 x 2 + 6 x − 1
Complete the square to find the
vertex:
f ( x) = −3( x 2 − 2 x + ) −1
_____
f ( x) = −3( x − 2 x + 1) − 1 − 3
2
f ( x) = −3( x − 1) 2 − 4
1
Completing the square, continued
• The vertex is (1 , 2)
• The quadratic function opens down since
the coefficient of the x squared term is
negative (-3) .
• Set f(x) = 0
0 = −3 x 2 + 6 x − 1
• Use the quadratic formula:
X=
−b ± b 2 − 4ac
2a
2(−3) −6
f ( x) = −3x 2 + 6 x − 1
•
f ( x) = −3(0) 2 + 6(0) − 1
Find the y-intercept : Let x = 0 and solve for y:
We have (0, -1)
2
Generalization
• Summary:
f ( x ) = a ( x − h) 2 + k
• where a is not equal to zero.
Graph of f is a parabola:
if a > 0, the graph opens
upward
if a < 0 , the graph opens
downward.
Generalization, continued
• Vertex is (h , k)
• Line or axis of symmetry: x = h
• f(h) = k is the minimum if a > 0, otherwise, f(h) = k is
the maximum
• Domain : set of all real numbers
• Range: { y y ≤ k} if a < 0. If a > 0, the range is
{ y y ≥ k}
3
Solution
• Solution: Yield= number of peaches per
tree x number of trees
• Yield = 300 x 20 = 6000 ( currently)
• Plant one more tree: Yield = ( 300 – 1(10))
* ( 20 + 1) = 290 x 21 = 6090 peaches.
• Plant two more trees:
• Yield = ( 300 – 2(10)* ( 20 + 2) = 280 x 22
= 6160
Solution, continued
• Let x represent the number of additional
trees. Then Yield =( 300 – 10x) (20 + x)=
• Y(x)= −10 x 2 + 100 x + 6000
• To find the maximum yield, note that the Y(x) function is
a quadratic function opening downward. Hence, the
vertex of the function will be the maximum value of the
yield.
•
Solution, continued
• Complete the square to find the vertex of the parabola:
•
• Y(x) =
−10( x − 10 x + 25) + 6000 + 250
2
4
Solution,continued
• Y(x)= −10( x − 5) 2 + 6250