MAT 122 Homework 4 Solutions
Section 2.1, Problem 2 —
Part a: The value of f 0 (1950) is negative. Observe that the tangent line for the graph at that
point would appear to be a decreasing linear function, hence have negative slope. This means that
the number of farms is decreasing.
Part b: The tangent line at t = 1960 would appear to be decreasing more quickly than the tangent line at t = 1980. This means that the slope of the tangent line at t = 1960 is more negative,
so f 0 (1960) is more negative that f 0 (1980).
Section 2.1, Problem 10 —
Part a: A graph of g(t) = (0.8)t is shown below in blue, while the tangent line at t = 2 is
graphed in yellow. The slope of this tangent line is the value g 0 (2). Since this line has negative
slope, the value of g 0 (2) is negative.
Part b: We estimate g 0 (2) using the small interval [2, 2.01]. The average value of g over that
interval is
g(2.01) g(2)
avgROC =
⇡ 0.142653
2.01 2
We can therefore estimate that g 0 (2) = 0.142653. Other answers on this question are possible.
We will later show how this derivative can be computed exactly. This will give that
✓ ◆
16
5
0
g (2) =
ln
⇡ 0.142812
25
4
so we can see that our estimate is relatively accurate.
Section 2.1, Problem 16 —
Part a: The function appears to be increasing on the entire interval [4, 8], so we can guess that
f 0 (6) should be positive. This indicates that the percentage of household with cable TV is growing
in the year 1996.
Part b: To estimate the value of a derivative from a table, we take the average rate of change
on the smallest interval possible. Here, these intervals have length 2. To estimate the value of the
derivative at t = 2, we compute the average rate of change on the intervals [0, 2] and [2, 4]. These
1
are
61.5
2
63.4
avgROC for [2, 4] =
4
avgROC for [0, 2] =
59.0
= 1.25
0
61.5
= 0.95
2
Either of these could be an estimate for the value of f 0 (2), but we can get another estimate by
averaging them, to obtain the estimate
f 0 (2) ⇡
1.25 + .95
= 1.1
2
We repeat the same process for f 0 (10) —
67.8 67.4
= 0.2
10 8
68.9 67.8
avgROC for [10, 12] =
= 0.55
12 10
avgROC for [8, 10] =
Again either of these would be an acceptable estimate, but their average yields another estimate,
f 0 (10) ⇡
0.2 + 0.55
= 0.375
2
Any of these estimates has units of % of households/year. Each says that the percentage of
people with cable television is growing each year, although we can observe that the growth is slower
in the year 2000 (t = 10) than in the year 1992 (t = 2).
Section 2.1, Problem 18 —
Point A is a point on the graph of our function, with x-coordinate 4. Since we have that
f (4) = 25, the coordinates of point A are (4, 25).
Points B and C are points on the line tangent to the graph at x = 4. We know that this line
passes through the point (4, 25), and we also know that the slope of this line is 1.5, since f 0 (4) = 1.5.
Using point-slope form, the equation of this line is
y
25 = 1.5(x
4) , y = 1.5x + 19
Therefore to find the y-coordinates of each point, we need only plug the x-coordinate of each into the
equation of this line. Since point C has an x-coordinate of 3.9, it has y-coordinate (1.5)(3.9) + 19 =
24.85. A similar computation shows that the y-coordinate of B is (1.5)(4.2) + 19 = 25.3. Thus the
coordinates of B and C are (4.2, 25.3) and (3.9, 24.85) respectively.
2
Section 2.1, Problem 20 — An approximate reproduction of the graph is shown below, which
we’ll use to answer the questions.
Figure 1: Graph for Problem 2.1.20
Part a: The value of f (4) is the y-value of the graph of f (x) at x = 4. The yellow dotted line
below is the constant function with value equal to f (4).
Figure 2: Graph for Part A
Part b: The di↵erence f (4) f (2) is the di↵erence between the y-values of the points on the
graph at x = 2 and x = 4. It is represented by the distance between the two dotted lines on the
graph below.
Figure 3: Graph for Part B
f (5) f (2)
is the slope of the secant line that passes through the graph at
5 2
x = 2 and x = 5, shown below on the graph in yellow.
Part d: The value of f 0 (3) is the slope of the line tangent to the graph at the point x = 3, shown
below on the graph in yellow.
Part c: The value
3
Figure 4: Graph for Part C
Figure 5: Graph for Part D
Section 2.2, Problem 10 —
Part a: f (x3 ) is the largest, as this is the place where the function’s value is the largest.
Part b: f (x4 ) is the smallest, as this is the place where the function’s value is least.
Part c: f 0 (x5 ) is the largest, as this is the place where the slope of the tangent line to the graph
is the greatest.
Part d: f 0 (x3 ) is the smallest, as this is the place where the slope of the tangent line to the
graph is the least. Notice that this is the only one of the given points where the slope of the tangent
line is negative.
4
Section 2.2, Problem 14 — One possible graph is shown below. Notice that the original
graph is increasing on the interval ( 1, 1) and decreasing on the interval (1, 1), with a “peak” at
x = 1 that has a horizontal tangent line. Therefore the derivative function should be positive on
the interval ( 1, 1), have a horizontal intercept at x = 1, and be negative on the interval (1, 1).
Figure 6: Graph for Problem 14
Section 2.2, Problem 16 — One possible graph is shown below. Notice that the original
graph in always increasing, so the derivative should be always 0. The original graph also has a
horizontal tangent line at x = 0, so the derivative function should have a value of 0 at x = 0.
Figure 7: Graph for Problem 16
Section 2.2, Problem 30 —
Part a: That f 0 (1) > 0 means that the function should have a tangent line at x = 1 with positive
slope. We see that the two options for this are graphs II and III. In graph III we observe that f 0 (x)
is decreasing. As we move right along the graph, the slopes of the tangent lines get smaller, with
f 0 (2) = 0 and negative values for f 0 (x) when x > 2. Therefore the correct graph is Graph III.
Part b: Again we have that the graph must be either Graph II or III. Since Graph III was the
answer to the previous question, we know that the correct graph is Graph II.
Part c: By process of elimination, we must have that this is either graph I or IV. Note that this
matches with the fact that f 0 (0) < 0, as both graphs I and IV have tangent lines at x = 1 with
negative slope. We observe that as we move right along the graph in Graph IV, the slopes of the
tangent lines are decreasing (becoming more negative), so the correct graph is Graph IV.
Part d: Process of elimination gives that the correct graph is Graph I. We can also observe,
though, that Graph I matches the given description.
Section 2.3, Problem 2 —
The derivative can be written as
dD
, and has units f eet/minute.
dt
5