Honors 2201-4
Homework 4 Solutions
Fall 2003
Section 2.1
(12, 18, 24) The equation for the line is
!
−7 − (−3)
y − (−7) =
(x − 3)
3−1
−4
y+7=
(x − 3)
2
y = −2x + 6 − 7 = −2x − 1
The difference quotient is calculated for the points (0, −1) and (−1, 1) as
∆y
−1 − 1
−2
=
=
= −2
∆x
0 − (−1)
1
and for the points (1, −3) and (−2, 3) as
∆y
−3 − 3
−6
=
=
= −2
∆x
1 − (−2)
3
(45) The average rates of change are
1−0
=1
[0, 1] :
1−0
[1, 2] :
[2, 3] :
23 − 13
8−1
=
=7
2−1
1
33 − 23
27 − 8
=
= 19
3−2
1
43 − 33
64 − 27
=
= 37
4−3
1
To find the pattern, we calculate the average rate of change symbolically.
[3, 4] :
[n, n + 1] :
n3 + 3n2 + 3n + 1 − n3
(n + 1)3 − n3
=
= 3n2 + 3n + 1
(n + 1) − n
1
To check this formula, lets look at the intervals [0, 1] and [3, 4]. When n = 0,
the formula predicts an average rate of change of 0 + 0 + 1 = 1, as calculated
previously. When n = 3, the formula predicts an average rate of change of
3(32 ) + 3(3) + 1 = 27 + 9 + 1 = 37, as calculated previously.
1