1. The table shows world average daily oil consumption from 1985 to 2010 measured in
thousands of barrels per day.
(a) Compute and interpret the average rate of change from 1990 to 2005. What are
the units?
(b) Estimate the instantaneous rate of change in 2000 by taking the average of two
average rates of change. What are its units?
Solution:
(a) The average rate of change from 1990 to 2005 is given by
84, 077 − 66, 533
17, 544
=
= 1169.6.
20 − 5
15
Thus, between the years from 1990 to 2005 oil consumption rose, on average, by
1169.6 thousands of barrels of oil per day per year. The unit is: thousands of
barrels of oil per day per year.
(b) To estimate the instantaneous rate of change in 2000, we will use the average of
the average rates of change from 1995 to 2000, and 2000 to 2005.
The average rate of change from 1995 to 2000 is given by
76, 784 − 70, 099
6685
=
= 1337
15 − 10
5
The average rate of change from 2000 to 2005 is given by
7293
84, 077 − 76, 784
=
= 1458.6
20 − 15
5
Taking the average of the two numbers, we get an estimate for the instantaneous
rate of change in 2000: 1397.8 thousands of barrels of oil per day per year.
2
2. For the function g whose graph is given, arrange the following numbers in increasing
order and explain your reasoning: 0, g 0 (−2), g 0 (0), g 0 (2), g 0 (4).
Solution:
At x = −2, x = 2 and x = 4, the slope of the tangent line to g is
positive. Moreover, the slope is steepest at x = −2, followed by x = 2 and then x = 4.
Thus, we have so far that 0 < g 0 (4) < g 0 (2) < g 0 (−2). Since the slope of the tangent
line at x = 0 is negative. We can conclude that g 0 (0) < 0 < g 0 (4) < g 0 (2) < g 0 (−2)
3
3. Sketch the graph of a function g for which:
g(0) = g(2) = g(4) = 0,
g 0 (1) = g 0 (3) = 0,
g 0 (0) = g 0 (4) = 1,
g 0 (2) = −1,
limx→5− g(x) = ∞ and limx→−1+ g(x) = −∞.
Solution:
4
4. Given f (x) =
Solution:
√
1 − 2x, find f 0 (a).
By definition, we have that
f (x) − f (a)
x→a
√ x−a √
1 − 2x − 1 − 2a
= lim
x→a
x−a
√
√
√
√
1 − 2x − 1 − 2a
1 − 2x + 1 − 2a
√
·√
= lim
x→a
x−a
1 − 2x + 1 − 2a
(1 − 2x) − (1 − 2a)
√
√
= lim
x→a (x − a) · ( 1 − 2x +
1 − 2a)
−2(x − a)
√
√
= lim
x→a (x − a) · ( 1 − 2x +
1 − 2a)
−2
√
= lim √
x→a
1 − 2x + 1 − 2a
−2
√
=√
1 − 2a + 1 − 2a
1
= −√
1 − 2a
f 0 (a) = lim
5