Section 2.1 – How Do We Measure Speed?
1.
(a) Given to the right is the graph of the
position of a runner as a function
of time. Use the graph to complete
each of the following.
d (feet)
40
S1
30
S2
20
T
10
1
Time Interval
2≤t≤5
2
3
4
t (seconds)
5
6
Average Velocity of Runner
The average velocity is
36 − 8
f (5) − f (2)
= 9.33 ft/sec.
=
3
5−2
Graphically, this average velocity is the slope of the line S1 that we have
drawn in above.
2 ≤ t ≤ 3.5
The average velocity is
f (3.5) − f (2)
18 − 8
=
= 6.67 ft/sec.
3.5 − 2
1.5
Graphically, this average velocity is the slope of the line S2 that we have
drawn in above.
2 ≤ t ≤ 2.5
The average velocity is
10.5 − 8
f (2.5) − f (2)
= 5 ft/sec.
=
0.5
2.5 − 2
(b) Estimate the instantaneous velocity of the runner at t = 2 seconds.
The instantaneous velocity of the runner at t = 2 seconds is the number that the average
velocities in the table from part (a) above approach as the length of our time
intervals approach 0. Graphically, the instantaneous velocity is the slope of the line
that we would see if we zoomed in on the portion of our position graph near t = 2 until
it looked like a straight line. This line has the same slope as the tangent line, T,
that we drew in on the above graph, so the instantaneous velocity at t = 2 is the slope
of T. Since the points (2, 8) and (4, 15) appear close to the line T, we can approximate
the slope of T as follows:
15 − 8
Slope of T =
= 3.5
4−2
Therefore, the instantaneous velocity of the runner at t = 2 is about 3.5 ft/sec.
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
1
2. For the function shown to the right, answer the following
questions:
4
A
E
3
(a) At what points is the slope of the curve positive?
The slope of the curve is positive at A,
D, and E because the slopes of the line
segments that we drew in to the right are
positive at these three points.
2
B
D
1
C
(b) At what points is the slope of the curve negative?
1
2
3
4
5
6
The slope of the curve is negative at B and
C because the slopes of the line segments
that we drew in to the right are negative
at these three points.
(c) Rank the slopes at the 5 points in order from smallest to largest.
Slope at B < Slope at C < Slope at A < Slope at E < Slope at D
3. In a time of t seconds, a particle moves a distance of s meters from its starting point, where s = f (t) = t2 + 1.
(a) Find the average velocity between t = 2 and t = 2 + h if h = 0.1, h = 0.01, and h = 0.001. (That is,
compute the average velocity over 3 different time intervals).
f (2.1) − f (2)
0.1
f (2.01) − f (2)
0.01
f (2.001) − f (2)
0.001
h = 0.1 :
h = 0.01 :
h = 0.001 :
=
=
=
5.41 − 5
0.1
5.0401 − 5
0.01
5.004001 − 5
0.001
=
4.1 m/sec
=
4.01 m/sec
=
4.001 m/sec
(b) Now, give your best estimate of the instantaneous velocity of the particle at t = 2.
As h approaches 0 in the calculations from part (a), it appears that the average
velocities of the particle are approaching 4 meters per second. Therefore, our best
estimate of the instantaneous velocity of the particle at t = 2 is 4 meters per second.
4. The position of a car traveling along a straight east/west highway at various times is shown in the table
below. Positive values of d indicate that the car is east of its starting point, while negative values of d
indicate that the car is west of its starting point.
t (hours)
d (miles)
1
40
2
-10
3
20
4
90
5
-50
Calculate the average velocity of the car on the following two time intervals: (a) between 1 and 2 hours, (b)
between 2 and 4 hours. What does a positive velocity mean? What does a negative velocity mean?
Δd
−10 − 40
=
= −50 miles per hour.
Δt
2−1
Δd
90 − (−10)
(b)
=
= 50 miles per hour.
Δt
4−2
(a)
Positive velocities indicate eastward travel, while negative velocities indicate westward
travel.
2
Developed by Jerry Morris
Section 2.1 – How Do We Measure Speed - Angry Mike
Officer Tom “Tommy Boy” Hopkins and his long-time adversary and evil brother Mike, known throughout the
county as “Angry Mike,” are having another one of their disputes. Angry Mike decides to show off his love for
lawlessness by speeding past the police station, but unbeknownst to him, Tommy Boy has set up an elaborate
system of invisible laser beams and timing devices along the road passing the police station. As a result, Tommy
Boy and his colleagues are able to construct a graph of Angry Mike’s distance traveled, x, as a function of time, t
(shown on the following page).
Alas, the case of Hopkins vs. Hopkins goes to trial, and you’re playing the part of the prosecution’s star witness.
1. Compute the average velocities (in miles per hour) on each of the following intervals.
(a) 0 seconds ≤ t ≤ 2 seconds
From the graph, we see that Angry Mike travels about 115 feet in during these 2
seconds. Therefore, his average velocity is given by
Δx
115
=
= 57.5 feet per second,
Δt
2
which, geometrically, is the slope of the line segment on the provided graph that goes
through the points at t = 0 and t = 2 (see the graph on the following page). Now, we
want to convert this number to a velocity in miles per hour, so we have
Average Velocity =
57.5 ft
1 mi
3600 sec
×
×
= 39.2 miles per hour,
1 sec
5280 ft
1 hr
which is our final answer. Note that to obtain this answer, we just multiplied the
average velocity, in feet per second, times the conversion factor 3600/5280.
(b) 2 seconds ≤ t ≤ 4 seconds
Proceeding as above, we have
Δx 3600
310 − 115 3600
×
=
×
= 66.5,
Δt
5280
2
5280
so the average velocity is 66.5 miles per hour.
(c) 4 seconds ≤ t ≤ 6 seconds
Δx 3600
610 − 310 3600
×
=
×
= 102.3,
Δt
5280
2
5280
so the average velocity is 102.3 miles per hour.
(d) 6 seconds ≤ t ≤ 8 seconds
Δx 3600
910 − 610 3600
×
=
×
= 102.3,
Δt
5280
2
5280
so the average velocity is 102.3 miles per hour.
(e) 8 seconds ≤ t ≤ 10 seconds
Δx 3600
1090 − 910 3600
×
=
×
= 61.4,
Δt
5280
2
5280
so the average velocity is 61.4 miles per hour.
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
3
(f) 10 seconds ≤ t ≤ 12 seconds
Δx 3600
1190 − 1090 3600
×
=
×
= 34.1,
Δt
5280
2
5280
so the average velocity is 34.1 miles per hour.
2. In opening arguments, Angry Mike’s defense attorney looks at the data on the graph and says that his client
traveled less than 1190 feet in 12 seconds, giving a top possible velocity of under 68 miles per hour. He
therefore claims that his client was traveling cautiously below the speed limit for the entire 12-second period.
Is the attorney’s claim accurate? Explain, based on your own calculations.
The defense attorney’s statement that the client traveled 1190 feet in 12 seconds is
accurate, but this information means that Angry Mike’s average velocity over the 12-second
period is about 68 miles per hour; it does not mean that 68 miles per hour was Angry Mike’s
top speed over this interval. Indeed, we see from our calculations above that Angry Mike
was traveling at an average velocity of about 102 miles per hour between t = 4 and t = 8
seconds.
3. Estimate Angry Mike’s maximum velocity during the 12-second interval. Explain how you got this number.
How can you tell from the graph where Angry Mike is traveling the fastest?
We can graphically determine where Angry Mike is traveling the fastest by finding the point
on the provided graph where the slope is the steepest. It appears that the graph is
steepest around t = 6 seconds, and we can approximate the slope here by drawing in the
tangent line to the graph at t = 6 and estimating its slope (see the graph on the following
page). Since this line appears to go through the points (5, 450) and (8, 930), we estimate
Angry Mike’s maximum velocity as follows:
Δx 3600
930 − 450 3600
×
=
×
= 109.1 miles per hour.
Δt
5280
8−5
5280
4. The defense attorney, in response to your testimony, claims that since all your velocity calculations are
averages over time intervals, rather than instantaneous velocities, that you can’t prove that Angry Mike was
speeding at any one given time. How would you respond to this statement? How would you justify your
position to the jury?
First, we could point out that our estimate of Angry Mike’s maximum velocity from question
3 above is a very accurate estimate of his instantaneous velocity at t = 6 because we based
our calculation on the slope of the provided graph at just one point, as opposed to
previous average velocity calculations. Furthermore, even if we had only presented the
average velocity data from question 1, the fact that Angry Mike was averaging 102.3 miles
per hour between t = 4 and t = 6 seconds indicates that either he was traveling at a constant
speed of 102.3 miles per hour for this entire interval, or he was traveling above 102.3
miles per hour for at least part of this time interval; otherwise, it would be impossible
for him to average a speed of 102.3 miles per hour! Finally, we could point out that,
because the provided graph is concave down on the interval 6 ≤ t ≤ 8, the average velocity of
102.3 miles per hour is actually an underestimate of his instantaneous velocity at any point
in the interval; therefore, it is indisputable that Angry Mike was traveling at least 102.3
miles per hour at every time between 6 and 8 seconds.
5. Finally, suppose (just this once!) that you’re working for the defense attorney’s team. As part of the closing
argument, to show what a decent, law-abiding, courteous, and cautious citizen Angry Mike is, give your best
estimate (as low as you can make it from the data) of his velocity (in miles per hour) at t = 12 seconds.
Explain why you think your answer is a good estimate of his velocity right at t = 12 seconds.
4
Developed by Jerry Morris
To estimate Angry Mike’s velocity at t = 12 seconds, we need to estimate the slope of the
given graph right at t = 12 seconds. To do this, we draw in a tangent line at t = 12 (see
the graph on the following page) and estimate its slope. Since this tangent line appears
to go through the points (9, 1060) and (12, 1190), we can estimate Angry Mike’s velocity at
t = 12 seconds as follows:
Δx 3600
1190 − 1060 3600
×
=
×
= 29.5 miles per hour.
Δt
5280
12 − 9
5280
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
5
x
1200
1000
800
600
400
200
t
2
6
4
6
8
10
12
Developed by Jerry Morris
Section 2.2 – The Derivative at a Point
f(a+h)
S
f(a)
a
a
a+h
Note: The slope of the secant
line, S, approaches the slope
of the tangent line as h
approaches 0.
f(a+h)
S
f(a)
a
a+h
Definition. The average rate of change of f over the interval a to a + h is given by
f (a + h) − f (a)
h
Note: Graphically, the average rate of change is the slope of the secant line, S.
Definition. The instantaneous rate of change of f at a, called the derivative of f at a, is given by
f (a + h) − f (a)
h→0
h
f ' (a) = lim
Note: Graphically, the instantaneous rate of change of f at a is the slope of the curve f
at a, or, equivalently, the slope of the tangent line to f at x = a.
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
7
1. Given to the right is the graph of a function f.
f(x)
(a) Rank the following quantities in order from smallest
to largest: f (0), f (a), f (b), f (1), f (c), f (d)
f (a) < f (0) < f (b) < f (1) < f (c) < f (d)
(b) Rank the following quantities in order from smallest
to largest: f ' (0), f ' (a), f ' (b), f ' (1), f ' (c), f ' (d)
a
b
1 c
d
f ' (0) < f ' (a) < f ' (d) < f ' (c) < f ' (1) < f ' (b)
2. Let f (x) = ln x. Estimate f ' (3) accurate to 2 decimal places.
Since
f (3 + h) − f (3)
ln(3 + h) − ln 3
= lim
,
h→0
h→0
h
h
f ' (3) = lim
we can use the following table to estimate the value of f ' (3).
h
ln(3 + h) − ln 3
h
0.1
0.01
0.001
0.3279
0.3328
0.3333
Therefore, f ' (3) = 0.33 to 2 decimal places.
3. Let f (x) = x2 − 3x − 2.
(a) Use algebra to find f ' (x) at x = 2.
f (2 + h) − f (2)
h→0
h
f ' (2) = lim
=
=
=
=
(2 + h)2 − 3(2 + h) − 2 − (22 − 3(2) − 2)
h→0
h
lim
lim
h→0
h(h + 1)
h
lim (h + 1)
h→0
1
Therefore, f ' (2) = 1.
(b) Find the equation of the tangent line to f at x = 2.
From part (a), the slope of the tangent line is f ' (2) = 1. One point on the tangent line
is (2, f (2)) = (2, 22 − 3(2) − 2) = (2, −4). Therefore, we have
y − (−4)
y+4
y
=
=
=
1(x − 2)
x−2
x − 6,
so y = x − 6 is the equation of the tangent line.
8
Developed by Jerry Morris
Section 2.3 – The Derivative Function
1. To the right you are given the graph of a function f (x).
With a straightedge, draw in tangent lines to f (x) at
every half unit and estimate their slope from the grid.
Use your slopes to draw a graph of f ' (x) on the same
set of axes.
After drawing in the tangent line segments
to the right and estimating their slopes, we
arrive at the following approximate figures:
3
2
1
2
1
1
2
1
x
f ' (x)
-1.5
-1.9
-1
-0.8
-0.5
-0.25
0
0
0.5
0.25
1
0.8
1.5
1.9
2
3
2. Given the graph of f (x) below, sketch a rough graph of f ' (x).
’
Notes:f ' (x) = 0 (i.e. f ' (x) crosses the x-axis) when the slope of f at x equals zero.
f ' (x) > 0 (i.e. f ' (x) is above the x-axis) when f is increasing.
f ' (x) < 0 (i.e. f ' (x) is below the x-axis) when f is decreasing.
If f has a cusp (sharp point) at x, then f ' (x) is undefined.
3. Given below is the graph of a function y = f (x). Sketch a rough graph of f ' (x) and of f '' (x) on the same set
of axes. Use different colors so you can tell the functions apart.
’’
’
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
9
4. Given the graph of f (x) below, sketch an accurate graph of f ' (x).
Since f is linear with a slope of −3/2 when
x < 0, this means that f ' (x) = −3/2 whenever
x < 0. Similarly, since f is linear with a slope
of 1 when x > 0, f ' (x) = 1 whenever x > 0. Since f
has a sharp point at x = 0, we know that f ' (0) is
undefined.
1
1
’
5. Use algebra (that is, use the limit definition of the derivative) to find a formula for f ' (x) for each of the
following functions.
(a) f (x) = x2
2
2x−2(x+h)
x(x+h)
2
−
f (x + h) − f (x)
f (x) = lim
= lim x+h x
h→0
h→0
h
h
'
=
lim
h
h→0
=
=
=
lim
h→0
lim
−2h
1
·
x(x + h) h
−2
h→0 x(x + h)
2
− 2,
x
so f ' (x) = −2/x2 .
√
(b) f (x) = x
f (x + h) − f (x)
= lim
f (x) = lim
h→0
h→0
h
'
√
√
x+h− x
h
=
=
=
=
√
√
√
√ x+h− x
x+h+ x
lim
·√
√
h→0
h
x+h+ x
lim
x+h−x
√
√
x + h + x)
lim
h
√
√
x + h + x)
h→0 h(
h→0 h(
1
√ ,
x+ x
√
1
so f ' (x) = 2√
.
x
10
Developed by Jerry Morris
Section 2.4 – Interpretations of the Derivative
1. Let f (p) represent the daily demand for San Francisco ’49ers T-shirts when the price for a shirt is p dollars.
In other words, f (p) gives the number of shirts purchased daily if the selling price is p dollars.
(a) Is f increasing or decreasing?
f(p) (shirts)
f is decreasing.
Slope units: shirts
dollar
(b) What are the units of p, f (p), and f ' (p)?
p has units of dollars, and f (p) has units of
shirts. Since f ' (p) represents the rate of
change of f (p) with respect to p, its units are
shirts per dollar.
p (dollars)
20
(c) Explain, in terms of shirts and dollars, the practical meaning of the following:
i. f (20) = 150
This states that if the price for a shirt is $20, then 150 shirts will be bought
daily.
ii. f ' (20) = −5
This states that if the price for a shirt is $20, then the daily demand for shirts
will decrease by about 5 shirts for each additional dollar that is charged for the
shirt.
iii. f (30)
This represents the number of shirts that will be bought in a day if the price for
a shirt is $30.
(d) Let d represent demand. Then d = f (p), so the function f takes price as an input and gives demand as
an output. On the other hand, the inverse function f −1 takes demand as an input and gives price as an
output, so f −1 (d) = p.
(e) Give practical interpretations of f (25) and f −1 (25).
f (25) represents the number of shirts bought daily if the selling price is $25.
f −1 (25) represents the selling price of a shirt that would result in 25 shirts being
bought daily.
2. (Taken from Hughes-Hallett, et. al.) If t is the number of years since 1993, the population, P, of China, in
billions, can be approximated by the function
P = f (t) = 1.15(1.014)t .
(a) Calculate and interpret f (6) in the context of this problem.
f (6) = 1.25, which indicates that the population of China was 1.25 billion in 1999.
(b) Use the table method to estimate dP
dt at t = 6, and give an interpretation of this number in the context
of this problem.
1.15(1.014)6+h − 1.15(1.014)6
h→0
h
P ' (6) = lim
h
1.15(1.014)6+h − 1.15(1.014)6
h
0.1
0.01
0.001
0.01739
0.01738
0.01738
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
From the calculations to the left,
P ' (6) equals about 0.01738, which
means that the population of China
is increasing by about 17.4 million
people per year in 1999.
11
3. Between noon and 6 p.m., the temperature in a town rises continually, but rises at its quickest around 3 p.m.,
and slowest around noon and 6 p.m.
(a) Sketch a possible graph of H = f (t), where H is the temperature in the town (in degrees Fahrenheit)
and t represents the time (in hours) after 12:00 noon.
H (deg F.)
f
3
6
t (hours)
(b) Explain, in terms of degrees and hours, what each of the following represents:
i. f ' (2)
iii. f (4) = 40
This represents the rate at which the
temperature increases, in degrees per hour,
at 2 p.m.
ii. f ' (3) = 7
This states that at 3 p.m., the temperature
is increasing at a rate of 7 degrees
Fahrenheit per hour.
This states that at 4 p.m., the temperature
in the town is 40 degrees Fahrenheit.
iv. f ' (4) = 1
This states that at 4 p.m., the temperature
in the town is increasing at a rate of 1
degree Fahrenheit per hour.
(c) Use the statements given in parts (iii) and (iv) from above to estimate the temperature in the town at
5:30 p.m. Is the actual temperature higher or lower than the estimate?
Estimated Temperature at 5:30 p.m.
= 40◦ F + (1
◦
F
) (1.5 hours) = 41.5◦ F
hr
This estimate is higher than the actual temperature at 5 : 30 p.m. because this
calculation assumes that the temperature continues to increase at the same rate of 1
degree Fahrenheit per hour between 4 and 5 : 30 p.m. We can see from our graph in part
(a), however, that f gets less and less steep between 4 and 5 : 30 p.m.
12
Developed by Jerry Morris
Section 2.5 – The Second Derivative
A function f is said to be...
1. increasing on an interval if its graph rises from left to right on that interval.
2. decreasing on an interval if its graph falls from left to right on that interval.
3. concave up on an interval if its graph is shaped like part, or all, of a right-side up bowl on that interval.
4. concave down on an interval if its graph is shaped like part, or all, of an upside down bowl on that
interval.
Example 1. Given below is the graph of a
function f. Estimate the intervals on which f is
increasing, decreasing, concave up, and concave
down.
8
f is increasing for 0 < x < 4.5 and 7.5 < x < 10.
f is decreasing for 4.5 < x < 7.5.
f is concave up for 0 < x < 2.5 and 6.2 < x < 8.5.
f is concave down for 2.5 < x < 6.2 and 8.5 < x <
10.
6
4
2
2
4
6
8
10
Example 2. Consider the four functions given below.
increasing
concave down
f is increasing, so f ' > 0
f ' is decreasing, so f '' < 0
decreasing
concave up
f is decreasing, so f ' < 0
f ' is increasing, so f '' > 0
increasing
concave up
f is increasing, so f ' > 0
f ' is increasing, so f '' > 0
decreasing
concave down
f is decreasing, so f ' < 0
f ' is decreasing, so f '' < 0
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
13
Summary of Key Facts. Assume that f is a function such that f ' and f '' both exist.
1. If f is increasing on an interval, then f ' (x) > 0 on that interval.
2. If f is decreasing on an interval, then f ' (x) < 0 on that interval.
3. If f is concave up on an interval, then f '' (x) > 0 on that interval.
4. If f is concave down on an interval, then f '' (x) < 0 on that interval.
Practice Problems
1. Given below are the graphs of three functions: f, g, and h.
f(x)
g(x)
2
2
2
−2
4
6
2
4
h(x)
2
6
2
−2
4
6
−2
Use the graphs to decide whether each of the quantities that follow are positive, negative, or zero.
(a) f ' (1) > 0
(b) f '' (1) < 0
(c) f ' (2) < 0
(d) f '' (2) < 0
(e) f ' (5) > 0
(f) f '' (5) > 0
2. A South Dakota driver cruising along I-90 speeds
up, sees a highway patrol car, and then begins to
slow down. A graph of her displacement as a function of time is shown to the right. Note. Positive
values of s indicate that the driver is east of her
starting point, and negative values of s indicate that
she is west of her starting point.
(g) g ' (3) < 0
(h) g '' (3) = 0
(i) h' (3) = 0
s (feet)
300
(j) h'' (3) = 0
(k) f (1) > 0
(l) h(1) < 0
The function s=f(t)
200
100
t (seconds)
1
2
3
4
5
(a) On what approximate intervals is f ' (t) positive? negative? Interpret the meaning of these intervals in
the context of this problem.
f ' (t) > 0 for 0 < t < 5, meaning that she is driving east for the entire 5 seconds (i.e.,
her easterly displacement increases over the entire 5-second interval).
(b) On what approximate intervals is f '' (t) positive? negative? Interpret the meaning of these intervals in
the context of this problem.
f '' (t) > 0 for 0 < t < 2.5, which indicates that the driver’s eastward velocity is
increasing. On the other hand, f '' (t) < 0 for 2.5 < t < 5, which indicates that the
driver’s eastward velocity is decreasing.
14
Developed by Jerry Morris
Section 2.5 – Derivatives
In each of the following situations, sketch the graph of a function f (t) that has the indicated properties.
f (t) is increasing
f (t) > 0
f ' (t) is increasing
f (t) > 0
f (t) is decreasing
f '' (t) > 0
f (t) is increasing
f '' (t) < 0
f (t) < 0
f '' (t) < 0
f ' (t) > 0
f (t) < 0
f (t) < 0
f (t) is decreasing
f ' (t) < 0
f ' (t) is increasing
f (t) > 0
f ' (t) < 0
f ' (t) < 0
f '' (t) < 0
f (t) > 0
f (t) is increasing
f (t) > 0
f '' (t) > 0
f ' (t) is increasing
f (t) is decreasing
f (t) < 0
f (t) > 0
f (t) is decreasing
f ' (t) is decreasing
f (t) < 0
f (t) is increasing
f (t) is increasing
f '' (t) < 0
f (t) < 0
f (t) is decreasing
f (t) > 0
f (t) is increasing
f '' (t) < 0
f ' (t) > 0
f (t) > 0
f '' (t) > 0
f ' (t) < 0
f '' (t) > 0
f (t) > 0
f (t) < 0
f '' (t) > 0
f ' (t) < 0
'
'
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
15
16
f ' (t) is decreasing
f (t) < 0
f ' (t) > 0
f ' (t) is increasing
f ' (t) > 0
f (t) > 0
f (t) > 0
f '' (t) < 0
f (t) is decreasing
f (t) is decreasing
f ' (t) is decreasing
f (t) < 0
f '' (t) < 0
f (t) > 0
f ' (t) > 0
f ' (t) > 0
f (t) < 0
f ' (t) is increasing
f (t) < 0
f ' (t) < 0
f '' (t) < 0
f ' (t) is increasing
f ' (t) < 0
f (t) < 0
f (t) < 0
f ' (t) is decreasing
f (t) is increasing
f ' (t) is decreasing
f (t) > 0
f (t) is increasing
f (t) is decreasing
f '' (t) > 0
f (t) < 0
f (t) is increasing
f (t) < 0
f '' (t) > 0
f '' (t) > 0
f ' (t) > 0
f (t) < 0
f (t) > 0
f (t) is decreasing
f ' (t) > 0
f ' (t) < 0
f (t) is decreasing
f (t) > 0
f (t) > 0
f ' (t) is increasing
f (t) is decreasing
'
'
Developed by Jerry Morris
Section 2.6 – Differentiability
16
16
f(x)
14
12
12
10
10
8
8
6
6
4
4
2
2
2
4
f(x)
14
6
8
10
2
lim f (x) = 8
4
6
10
8
10
lim f (x) = 8
x→4
x→4
f (4) = 2
f (4) = 8
f is not continuous at x = 4.
f is continuous at x = 4.
16
16
g(x)
14
12
10
10
8
8
6
6
4
4
2
2
4
g(x)
14
12
2
8
6
8
10
2
lim g(x) = 8
4
6
lim g(x) does not exist
x→4
x→4
g(4) is undefined
g(4) = 8
g is not continuous at x = 4.
g is not continuous at x = 4.
Theorem. The function f is continuous at x = c if f is defined at x = c and if
lim f (x) = f (c) .
x→c
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
17
1. Which of the following functions are continuous at x = 0?
f(x)
g(x)
h(x)
k(x)
m(x)
n(x)
f and k are not continuous at 0 because they are not defined there. m and n are not
continuous at 0 because the limits of these functions as x approaches 0 do not agree with
the respective function values at 0. To summarize, only g and h are continuous at 0.
2. Consider the function f (x) given below.
16
14
12
10
8
6
4
2
2
4
6
8
10
(a) At what values of x is f not continuous?
x = 4, x = 6, and x = 8.5
(b) At what values of x is f not differentiable?
x = 1.5, x = 4, x = 6, x = 7, and x = 8.5
18
Developed by Jerry Morris
Section 2.6 – Differentiability
1. A magnetic field, B, is given as a function of the distance, r, from the center of a wire as follows:
⎧
⎨ rr0 B0 for r ≤ r0
B=
⎩ r0
for r > r0
r B0
(a) Is B continuous at r0 ? Explain.
First, note that we have
lim B
r→r0+
=
lim
r
r
r→r0+
lim B
r→r0−
=
lim
r→r0−
0
B0 = B0
r
B0
r0
= B0 .
Therefore, lim B exists and equals B0 , the value of the magnetic field B at r0 , so B
r → r0
is continuous at r0 .
(b) Is B differentiable at r0 ? Explain.
Let B = f (r). Then
r0
B0 − B0
f (r0 + h) − f (r0 )
r0 − (r0 + h)
B0
= lim r0 +h
= B0 lim
=− ,
+
+
+
h
h
h(r0 + h)
r0
h→0
h→0
h→0
lim
and
f (r0 + h) − f (r0 )
= lim
−
h
h→0
h → 0−
lim
r0 +h
r0 B 0 − B 0
h
= B0 lim
h → 0−
r0 + h − r 0
B0
=
,
hr0
r0
so because the above two limits are not equal, we conclude that f ' (r0 ) does not exist.
Therefore, B is not differentiable at r0 .
2. Sketch the graph of y = f (x) if f has the following properties:
y
4
• f (x) is continuous everywhere except at 3
3
• f (x) has a vertical tangent line at 2
2
• f (x) is not differentiable at 3
1
• f '' (x) > 0 wherever it is defined
1
Activities to accompany Calculus, Hughes-Hallett et al, Wiley, 2020
2
3
4
x
19
3. Find the intersection point of the tangent line to y = xx at 1.1 and the x-axis.
Let f (x) = xx . We begin by writing
�
x
f (x) = xx = eln x = ex ln x ,
so, by the Chain Rule,
f ' (x) =
d � x ln x e
=
dx
=
=
d
(x ln x)
dx
1
x ln x
e
x · + ln x
x
ex ln x ·
xx (1 + ln x).
Therefore, the slope of the tangent line to y = xx at 1.1 is f ' (1.1) = 1.11.1 (1 + ln 1.1), and a
point on the tangent line is (1.1, f (1.1)) = (1.1, 1.11.1 ). Substituting this information into
the point-slope form for the equation of a line, we see that the equation of the tangent
line is given by
y − 1.11.1
=
1.11.1 (1 + ln 1.1)(x − 1.1).
To find the x-coordinate of the point of intersection of this line with the x-axis, we
substitute 0 for y in the above equation and solve for x.
−1.11.1
=
1.11.1 (1 + ln 1.1)(x − 1.1)
−1
=
(1 + ln 1.1)(x − 1.1)
x
=
1.1 −
1
1 + ln 1.1
Therefore,
the point
of intersection of this tangent line with the x-axis is given by
1
1.1 −
, 0 , which is approximately equal to (0.187, 0).
1 + ln 1.1
20
Developed by Jerry Morris