Sullivan Algebra and
Trigonometry: Section 3.2
Objectives
• Find the Average Rate of Change of Function
• Use a Graph to Determine Where a Function Is
Increasing, Is Decreasing, or Is Constant
• Use a Graph to Locate Maxima and Minima
• Determine Even, Odd Functions from a Graph
• Identify Even, Odd Functions from the Equation
If c is in the domain of a function y = f(x), the
average rate of change of f between c and x is
defined as
y f ( x )  f (c)
average rate of change =

x
xc
This expression is also called the difference
quotient of f at c.
The average rate of change of a function can be
thought of as the average “slope” of the function,
the change is y (rise) over the change in x (run).
y = f(x)
(x, f(x))
Secant Line
f(x) - f(c)
(c, f(c))
x-c
2
s
(
t
)


16
t
 100t  6
Example: The function
gives the height (in feet) of a ball thrown straight up
as a function of time, t (in seconds).
a. Find the average rate of change of the height of
the ball between 1 and t seconds.
s s(t )  s(1)

, t 1
t
t 1
s(t )  16t  100t  6
2
s(1)  16(1)  100(1)  6  90
2
s(t )  s(1)  16t 2  100t  6  90

t 1
t 1
 16t 2  100t  84

t 1




 4 4t 2  25t  21
t 1


 4 4t 2  25t  21
t 1
 4(4t  21)(t  1)

t 1
 4(4t  21)
b. Using the result found in part a, find the
average rate of change of the height of the
ball between 1 and 2 seconds.
Average Rate of Change between 1 second
and t seconds is: -4(4t - 21)
If t = 2, the average rate of change between 1
second and 2 seconds is: -4(4(2) - 21) = 52
ft/second.
A function f is increasing on an open
interval I if, for any choice of x1 and x2 in I,
with x1 < x2, we have f(x1) < f(x2).
A function f is decreasing on an open
interval I if, for any choice of x1 and x2 in I,
with x1 < x2, we have f(x1) > f(x2).
A function f is constant on an open interval I
if, for any choice of x in I, the values of f(x)
are equal.
Determine where the following graph is
increasing, decreasing and constant.
y
4
(2, 3)
(4, 0)
0
(1, 0)
x
(0, -3)
-4
(7, -3)
(10, -3)
A function f has a local maximum at c if
there is an interval I containing c so that, for
all x in I, f(x) < f(c). We call f(c) a local
maximum of f.
A function f has a local minimum at c if
there is an interval I containing c so that, for
all x in I, f(x) > f(c). We call f(c) a local
minimum of f.
Referring to the previous example, find all
local maximums and minimums of the
function:
y
4
(2, 3)
(4, 0)
0
(0, -3)
-4
(1, 0)
x
(7, -3)
(10, -3)
A function f is even if for every number x in
its domain the number -x is also in the
domain and f(x) = f(-x).
A function is even if and only if its graph is
symmetric with respect to the y-axis.
A function f is odd if for every number x in
its domain the number -x is also in the
domain and -f(x) = f(-x).
A function is odd if and only if its graph
is symmetric with respect to the origin.
Determine whether each of the following functions
is even, odd, or neither. Then determine whether
the graph is symmetric with respect to the y-axis or
with respect to the origin.
a.)
g ( z)   z 2  2
g (  z)  (  z)  2  z  2
2
2
g ( z)  g (  z)
Even function, graph symmetric with
respect to the y-axis.
b.)
f ( x)  4 x 5  3 x
f (  x )  4(  x )5  3(  x )  4 x 5  3x
f ( x)  f ( x)
Not an even function.
 f ( x )   ( 4 x 5  3x )  4 x 5  3x
f (  x )  4(  x )5  3(  x )  4 x 5  3x
f ( x)   f ( x)
Odd function, and the graph is
symmetric with respect to the origin.