3.3. Rates of Change
f(x) = 10 + 18x + 9x2 – 2x3
250
Average rate of change between s and t
200
f (t ) − f ( s )
ARC =
t−s
150
100
50
0
If f is the position this is the average speed
If f(1)=50 and f(3)=174 miles then
ARC = 124/2 = 62 miles per hour
-4
-3
-2
-1
0
-50
1
2
3
4
5
6
7
-100
-150
f(1)=35, f(0)=10, ARC = 25
ARC = (1.888)/(0.1)= 18.88
12
35
11.888
30
11.5
25
11
20
10.5
15
10
0
0.2
0.4
0.6
0.8
1
10
0
ARC = (.1809)/(.01) = 18.09
10.2
0.02
0.04
0.06
0.08
0.1
ARC = (.018009)/(.001) = 18.009
10.02
10.1809
10.018009
10.16
10.016
10.12
10.012
10.08
10.008
10.04
10.004
10
10
0
0.002
0.004
0.006
0.008
0.01
0
0.0002
0.0004
0.0006
0.0008
0.001
f '(x) = limh→0
h
1
0.1
0.01
0.001
limit
f (x + h) − f (x)
h
ave rate of change
25
18.88
18.09
18.0009
18
Derivative of f
f '( x) = lim h→0
Is the Instantaneous rate of change
If f(x) = number on the odometer
f′(x) = number on the speedometer
Why is the derivative = 18?
f(x) = 10 + 18x + 9x2 – 2x3
When x is small x2 and x3 are much
smaller than x so
f(x) ≈ 10 + 18x
Equation of the tangent line
f ( x + h) − f ( x )
h
Computing f′(4), Figure 28
100
95
90
85
80
Slopes -23, -14, -7.52, -6
75
70
3.9
Why is f′(4) = -6?
f(x) = 10 + 18x + 9x2 – 2x3
Next time we will learn that
f′(x) = 0 + 18(1) + 9(2x) – 2(3x2)
So we have
f′(4) = 18 + 72 – 96 = -6
4
4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9
5
Next time
Section 3.4 and part of 4.1
Homework #1 due in section Feb 1 or 2
3.3: 5, 27; 3.4: 14. f′(x), f′(3), 18, 46 c,d
4.1: 4,9
Homework, slides are on course web page