1. 3.7: Rates of Change
© 2007 Roy L. Gover (www.mrgover.com)
Learning Goals:
•Find the average rate of change of a
function over an interval
•Represent the avg. rate of change
geometrically as the slope of a secant
line
•Use the difference quotient to find a
formula for the average rate of change of
a function

2. Average Rate of Change
Time t 0 1 2 3 3.5 4 4.5 5
Distance
d(t)
0 16 64 144 196 256 324 400
Average speed= Distance traveled
time interval
Find the average speed of the falling rock
from t=3 to t=4.5
average speed= d(4.5)- d(3)=
4.5-3
324-144= 180=120
1.5 1.5

3. Try This
Find the average speed of
the Falling rock
a.From t=1 to t=4 80 ft/se
b.From t=2 to t=4.5
104 ft/sec
Time t 0 1 2 3 3.5 4 4.5 5
Distance
d(t)
0 16 64 144 196 256 324 400

4. Definition
Let f be a function
The average rate of change of f(x) with
respect to x as x changes from a to b is
the value
Change in f(x) = f(b)-f(a)
Change in x b-a

5. Example
A balloon is being filled with water.
Its approximate volume in gallons is
V(x)= x³
55
Where x is the radius of the balloon in
inches. Find the average rate of change of
the volume of the balloon as the radius
increases from 5 to 10 inches

6. Solution
Change in volume= V(10)-V(5) =
Change in radius 10-5
18.18-2.27 = 15.91= 3.18 gallons per inch
10-5 5

7. Example
A small manufacturing company makes
specialty office desks. The cost (in
thousands of dollars) of producing x
desks is given by the function:
C(x) = .0009x³-.06x²+1.6x+5
Find the average rate of change of cost
a.From 0 to 10 desks
b.From 10 to 30 desks
c.From 30 to 50 desks

8. Solution
a.15.9-5 = 10.9 = 1.09
10 10
b. 23.3-15.9= 7.4 = .37
30-10 20
c. 47.5 – 23.3 = 24.2 =1.21
50-30 20

9. Geometric Interpretation of Average Rate of
Change
Let f be a function.
Average rage of change= f(b)-f(a) =Slope of secant line
of f from x=a to x=b b-a joining (a,f(a)) and
(b,f(b)) on the graph of f

10. The Difference Quotient
Average rate of change is often computed
over very small intervals, such as the rate
from 4 to 4.01. Since 4.01= 4+.01 both
cases are essentially the same: computing
the rate of change over the interval from 4
to 4+h for some small quantity h.
The average rate of change for any function
f over a interval from x to x+h is:

11. The Difference Quotient
Average Rate of Change=
f(b) –f(a) = f(x+h)-f(x)= f(x+h)-f(x)
b-a (x+h)-x h

12. Example
Find the difference quotient of V(x)= x³
55
And use it to find the average rate of
change of V as x changes from 8 to 8.01

13. Solution
V(x+h)-V(x)= (x+h) - x³ = 1 ((x+h)³ -x³
h 55 55 55
h h
= 1 (x³ +3x²h+3xh²+h³-x³)= 3x² +3xh +h²
55 55
To find the average rate of change from 8 to
8.01
3x² +3xh+ h² = 3 ˣ 8² +3 ˣ 8(.01) + (.01)² =
55 55
3.495