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Calculus I (part 3): Applications of Dierentiation
(by Evan Dummit, 2016, v. 2.05)
Contents
3 Applications of Dierentiation
3.1
3.2
1
Minimum and Maximum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
3.1.1
Absolute Minimum and Maximum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
3.1.2
Local Minimum and Maximum Values, Critical Numbers
. . . . . . . . . . . . . . . . . . . .
3
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
Increasing and Decreasing Functions
3.3
3.2.1
Rolle's Theorem and the Mean Value Theorem
. . . . . . . . . . . . . . . . . . . . . . . . . .
3.2.2
Classication of Increasing and Decreasing Behavior
3.2.3
The First Derivative Test
8
. . . . . . . . . . . . . . . . . . . . . . .
9
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
Concavity, Graphing With Calculus
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1
Concavity, Inection Points, and the Second Derivative
. . . . . . . . . . . . . . . . . . . . .
3.3.2
Geometric Properties of Concavity, the Second Derivative Test
3.3.3
Graphing Functions Using Calculus
11
11
. . . . . . . . . . . . . . . . .
14
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
3.4
L'Hôpital's Rule
3.5
Applied Optimization (Functions of One Variable)
3.6
Antiderivatives, Applications to Physics
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
3 Applications of Dierentiation
Derivatives have a wide variety of applications. We will begin by discussing two closely related, and fundamental,
uses of the rst derivative: that of nding the largest and smallest values attained by a dierentiable function,
and that of understanding where a function is increasing or decreasing. Along the way, we will prove a number
of fundamental results about derivatives, including Rolle's Theorem, the Mean Value Theorem, and the First
Derivative Test. We will then turn our attention to the second derivative and use it to study concavity, and discuss
how to draw accurate graphs of functions.
Afterward, we discuss some other applications: L'Hôpital's rule for evaluating indeterminate limits, applied optimization, and antiderivatives with applications to physics.
3.1
•
Minimum and Maximum Values
In this section, our goal is to describe how to use calculus to nd minimum and maximum values of functions.
3.1.1 Absolute Minimum and Maximum Values
•
•
We are interested in nding minimum and maximum values, so we must rst dene what precisely this means:
Denitions: Suppose that
x=d
x in I .
at
if
f (x) ≤ f (d)
f (x) is a function dened on an interval I . We say f has an absolute maximum on I
x in I . We say f has an absolute minimum on I at x = c if f (x) ≥ f (c) for all
for all
1
◦
Terminology: The plural of maximum is maxima and the plural of minimum is minima. The word
extremum is used to refer to a point that is either a minimum or a maximum (also called extreme
points). The plural of extremum is extrema.
◦
We note that an absolute maximum need not be unique: if
values of
◦
x in I ,
f
takes the same maximum value for several
all of them are considered absolute maxima of
f.
(The same holds for minimum values.)
We also note that the absolute minimum and maximum of a particular function on an interval will depend
on the interval being considered.
◦
Example:
[0, 3π], the function f (x) = sin(x) has an absolute maximum value of
x = π/2 and x = 5π/2, and has an absolute minimum value of −1 occurring
On the interval
occurring both at
1,
at
x = 3π/2.
◦
Example: On the interval
occurring at
•
x = π/6,
[0, π/6],
the function
It may seem obvious that, given an interval
maximum somewhere in
◦
I.
On the open interval
f (x) = sin(x) has an absolute maximum
0 occurring at x = 0.
value of
1/2,
and has an absolute minimum value of
I
and a function
f,
that
f
will possess an absolute minimum and
However, this is not true! Here are a few examples illustrating what can go wrong:
(0, 1),
the function
f (x) = x
has neither an absolute minimum nor an absolute
maximum: although the function takes any suciently small positive value, it does not attain the value
0 on the interval
(0, 1).
Similarly, although the function takes values arbitrarily close to 1, it does not
attain the value 1 on the interval
◦
On the innite interval
at
x = 1,
[0, ∞),
(0, 1).
the function
f (x) = 1/x has an absolute maximum value of x occurring
x → ∞, the function approaches the value 0, but there is
but has no absolute minimum. As
f (x) is actually equal to 0.

2

2x − x for 0 ≤ x < 1
◦ The function f (x) = 1/2
on the closed interval [0, 2] has the
for x = 1


2x − x2 for 1 < x ≤ 2
value of 0 occurring at x = 0 and x = 2, but has no absolute maximum because
attains the value 1 anywhere in the interval [0, 2].
no real number
x
for which
2
absolute minimum
the function never
◦
The function


1 − x
g(x) = 1


3−x
for
for
for
0≤x<1
x=1
1<x≤2
on the closed interval
[0, 2]
has no absolute minimum or
maximum on the interval, since it approaches the values 0 and 2 arbitrarily closely but never attains
them.
•
From these examples, it seems that in order to ensure that
an interval
I,
we should require
f
f
to be continuous and make
has an absolute minimum and maximum on
I
contain its endpoints (i.e., require
I
to be a
nite, closed interval). In fact, these two conditions suce:
•
Theorem (Extreme Value):
If
f (x)
is continuous on the closed interval
[a, b],
then it attains its absolute
maximum and absolute minimum on that interval. In other words, there exist real numbers
M (the maximum) such
f (c) = m and f (d) = M .
and
◦
that
m ≤ f (x) ≤ M
on
[a, b],
and real numbers
c
and
d
m (the minimum)
[a, b] such that
in
The technical details of the proof of this theorem are not terribly enlightening, and it relies on a technical
property of the real numbers known as the least upper bound axiom, so we will omit the details.
3.1.2 Local Minimum and Maximum Values, Critical Numbers
•
There is another avor of maximum / minimum point for us to analyze:
•
Denition: For any function
all
x
f (x)
some open interval containing
◦
and any value
in some open interval containing
c.
c, we say f has a local maximum at x = c if f (x) ≤ f (c) for
f has a local minimum at x = c if f (x) ≥ f (c) for all x in
We say
c.
In contrast to an absolute maximum, being a local maximum only requires that
Likewise, being a local minimum only requires that
◦
Example: The function
f (x) = x3 − 3x
f
has a local maximum at
3
be smaller nearby.
x = 1 and a local minimum at x = −1.
[−2, 2], because f takes larger negative
Neither point is an absolute minimum or maximum on the interval
and larger positive values elsewhere on the interval.
f
be larger nearby.
◦
Example: The function
g(x) = x + 1/x
has a local maximum at
x = −1 and
[−4, 4].
a local minimum at
x = 1.
Neither point is an absolute minimum or maximum on the interval
•
We can often visually identify the locations where a function seems to take local minimum and maximum
values by looking at a graph.
However, this procedure is not rigorous, and will not generally give exact
answers.
◦
Example:
x ≈ −1.4,
and a local minimum at roughly
minimum is at
◦
f (x) = x3 − 6x
x ≈ 1.4. (In fact the
By studying the graph, it appears that
√
x =
√
2
and the
x = − 2.)
g(x) = ln(2x)/x has a local maximum
x ≈ 1.35. (In fact, the exact location of the
Example: By studying the graph, it appears that the function
(and also absolute maximum) value occurring roughly at
x = e/2.)
local maximum is
•
has a local maximum at roughly
maximum is at
Notice, in the examples above, that the tangent line to the curve
y = f (x)
at a local minimum or maximum
is always horizontal. This is not an accident:
•
f (x)
Theorem (Fermat): If
is dierentiable at
x=c
and
f
has a local maximum or minimum at
x = c,
then
f 0 (c) = 0.
◦
Proof: Suppose rst that
f
has a local minimum at
x = c.
We will show that
f 0 (c)
is both at least zero
and at most zero.
◦
By denition, if
f (c) ≤ f (x)
◦
f
Since
f
has a local maximum at
for all
x
is dierentiable at
x = c,
one-sided limits are equal to
f 0 (c) = lim
then there is some open interval around
c
such that
the two-sided limit
0
f (x) − f (c)
x→c
x−c
lim
exists. So in particular, both of the
f (c).
f (x) − f (c)
◦
Now
◦
In the same way, we also have
, and since f (c) ≤ f (x) and c < x for all x in the limit, we see that f
x→c+
x−c
0
is equal to a limit of nonnegative numbers, so by the inequality rule for limits we obtain f (c) ≥ 0.
limit, we see that
obtain
f 0 (c)
f 0 (c) = lim
0
(c)
f (x) − f (c)
, and since f (c) ≤ f (x) and x < c for all x in the
x→c−
x−c
is equal to a limit of nonpositive numbers, so by the inequality rule for limits we
0
f (c) ≤ 0.
◦
Combining the two inequalities
◦
Finally, in the event
(noting that
•
x=c
in that interval.
g
f
f 0 (c) ≤ 0
and
has a local maximum at
has a maximum at
x = c)
f 0 (c) ≥ 0
shows that
f 0 (c) = 0
as claimed.
x = c, we may apply the argument above to g(x) = −f (x)
g 0 (c) = 0, and thus f 0 (c) = −g 0 (c) = 0 as well.
to conclude that
Note, importantly, that the converse of Fermat's Theorem is
not true ! Even
x = c.
if
f 0 (c) = 0,
that does not
necessarily mean that there is a local minimum or local maximum at
◦
Example: The function
for
f (x) = x3
has
f 0 (0) = 0,
f.
4
but 0 is neither a local maximum nor a local minimum
◦
minimum for
•
f (x) = (x − 1)5
Example: The function
has
f 0 (1) = 0,
but 1 is neither a local maximum nor a local
f.
Since we are interested in looking for minimum and maximum points, we give a name to these places where
a minimum or maximum could potentially occur:
•
f (x) is a
point (a, f (a))
x
Denition: A critical number of a function
value of
x=a
is a critical point.
◦
is a critical number, we say the
Example: For
because
f 0 (0) = 0,
f 0 (x) = 0
we say 0 is a critical number of
or
f,
f 0 (x)
and
is undened. If
(0, 0)
is a critical
f.
point for
◦
f (x) = x2 ,
for which
Notation: The terms critical point and critical value are often used interchangeably with critical
number.
We will reserve critical point to refer to a point
(x, y)
on a graph where
x
is a critical
number. The distinction is not particularly important, because the concepts are so closely related, but it
can initially be confusing to switch freely from discussing critical numbers (representing
x-coordinates)
to discussing critical points (representing points on graphs).
•
To emphasize, critical points are locations where
points are not
where
◦
f0
necessarily
f could
have a local minimum or maximum. However, critical
either a minimum or a maximum. It is also very important to include locations
is undened, in addition to places where
Example: The function
f (x) = x3
f0
is zero.
has a critical point at the origin since
f 0 (0) = 0.
But the critical point
is neither a local maximum nor a local minimum, as we noted earlier.
◦
point is a local (and
•
f (x) = |x| has a critical point at the origin since f 0 (0) is not dened.
absolute) minimum, since |x| ≥ 0 for all x.
Example: The function
Example: Find the critical numbers of
p0 (x) = 5x4 − 5
◦
Because
◦
We must solve
p(x) = x5 − 5x.
is always dened, we need only nd the values for which
5x4 − 5 = 0,
5(x4 − 1) = 0.
x = 1 and x = −1.
or equivalently
so we get the two (real) solutions
The critical
5
Factoring yields
p0 (x) = 0.
5(x − 1)(x + 1)(x2 + 1) = 0,
◦
•
Thus, the critical numbers of
p
are
Example: Find the critical numbers of
(
◦
Since
h(x) =
3x − 2
2 − 3x
for
for
x = −1, 1
.
h(x) = |3x − 2|.
x ≥ 2/3
,
x < 2/3
(
3
h (x) =
−3
0
we see that
for
for
x > 2/3
x < 2/3
and that
h0 (2/3)
is unde-
ned.
◦
•
Since
h0
is never zero, we conclude that
From our discussion above, if
f
◦
f
are each attained either at a critical number of
b.
f
on the interval, we need only make a list of the
◦
3x2 − 3 = 0,
x = 1.
To solve
◦
The only critical number inside the interval is
We compute
◦
2
[0, 2],
(occurring at
p(x) = x3 − 3x
on the interval
the interval. Since
3(x − 1)(x + 1) = 0,
a and
[0, 2].
p0 (x) = 3x2 − 3
is always
so the critical numbers are
x = −1
x = 1, so (including the two endpoints) we have 3 potential
and
p(2) = 2.
the absolute minimum of
p
is
−2
(occurring at
x = 1)
and the absolute maximum
x = 2).
Here is the graph of
y = p(x),
with the critical point and endpoints marked:
Example: Find the absolute minimum and maximum values of
◦
at the endpoints
0, 1, 2.
p(0) = 0, p(1) = −2,
Therefore, on
is
f
0
we can simply factor to get
locations for the min and max:
◦
p(x) inside
p (x) = 0.
First, we make a list of the critical numbers of
and
•
f
The absolute maximum is the largest value on the list, while the absolute minimum is the smallest.
dened, the only critical numbers occur when
◦
[a, b], the absolute minimum
f , or at one of the endpoints
at all of the critical numbers in the interval, along with the values of
Example: Find the absolute minimum and maximum values of
◦
h.
[a, b].
Therefore, to nd the minimum and maximum values of
values of
•
is the only critical number of
is any continuous function dened on the interval
and absolute maximum values of
of the interval
x = 2/3
First, we make a list of the critical numbers of
f (x)
always dened, the only critical numbers occur when
f (x) = x + 2 sin(x)
on the interval
inside the interval. Since
[0, 3π].
f 0 (x) = 1 + 2 cos(x)
is
0
f (x) = 0.
1
2π 4π 8π
, we obtain three critical numbers in [0, 3π]: x =
,
,
.
2
3 3 3
2π 4π 8π
◦ Including the two endpoints, we have 5 potential locations for the min and max: 0,
,
,
, 3π .
3 3 3
√
√
√
2π
4π
4π
8π
8π
◦ We compute f (0) = 0, f ( 2π
3 ) = 3 + 3 ≈ 3.826, f ( 3 ) = 3 − 3 ≈ 2.457, f ( 3 ) = 3 + 3 ≈ 10.110,
and f (3π) = 3π ≈ 9.424.
◦
Since
f 0 (x) = 0 is equivalent to cos(x) = −
6
◦
Therefore, on
is
◦
•
8π
+
3
√
3
[0, 3π],
(occurring at
Here is the graph of
f
the absolute minimum of
x=
y = f (x),
is
0
(occurring at
x = 0)
and the absolute maximum
8π
3 ).
with the critical point and endpoints marked:
We would also like to be able to decide when a critical number of
f
is a local minimum or local maximum.
In order to do this most eciently, however, we must discuss another topic rst.
3.2
•
Increasing and Decreasing Functions
In the previous section, we studied the behavior of
f
at locations where the derivative
We now turn our attention to studying the behavior of
f
at points where
f
0
f0
was equal to zero.
is not zero: specically, we will
discuss the ramications that the sign of the derivative (positive or negative) has on the behavior of
•
By denition, the value of the derivative
◦
0
f (c)
measures how fast the value of
f (x)
is changing when
f.
x = c.
f 0 (c) > 0
then, almost by denition, intuition suggests that f should be increasing
x a small amount from c, f should take a larger value, and if we decrease x
slightly, then the value of f should also decrease.
◦ Inversely, if f 0 (c) < 0, f should be decreasing near x = c: if we increase x a small amount from c, f
should decrease, and if we decrease x, then f should increase.
In particular, if
near
•
x = c:
if we increase
In order to justify this intuitive sense rigorously we need to appeal to a few theorems, but rst we need to
give a better denition of what it means for a function to be increasing or decreasing:
•
Denitions: If
I.
We say
◦
f
f
is a function dened on an interval
is decreasing on
Example: The function
I
if
f (a) > f (b)
f (x) = x2
for all
I , we f is increasing
a < b in I .
is decreasing on the interval
on
[−1, 0]
I
if
f (a) < f (b)
for all
a<b
in
and increasing on the interval
[0, 1].
◦
Example: The function
f (x) = sin(x) is increasing on [0, π/2] and [3π/2, 2π] and decreasing on [π/2, 3π/2].
7
3.2.1 Rolle's Theorem and the Mean Value Theorem
•
We begin our analysis of increasing and decreasing functions with the following theorem, which will initially
seem to be completely unrelated:
•
Theorem (Rolle's Theorem): If
satises
◦
f (a) = f (b),
f (x)
is a continuous function on
then there exists some point
x = c, then
value of c.
Corollary: If
◦
f (a) = f (b)
◦
the only possibility is for
in the interval we want, and
f
Proof: If
(a, b)
•
c
(a, b)
and
x-coordinate
f 0 (c) = 0
of the minimum as
c.
Now if both the absolute maximum and minimum of the function
point
dierentiable on
by Fermat's Theorem we obtain that
Similarly, if the minimum is not at an endpoint, then we can use the
because
•
[a, b] which is
f 0 (c) = 0.
for which
f (x) will attain its absolute maximum and absolute
If the maximum is not at an endpoint, say at
the value of
◦
(a, b)
[a, b].
and so we can take this maximum as the
◦
in
Proof: By the Extreme Value Theorem, we know that
minimum somewhere on the interval
◦
c
f 0 (c)
f (x)
f (x)
will be zero.
is dierentiable everywhere, then between any two zeroes of
f (a) = 0 and f (b) = 0,
f 0 (c) = 0.
occur at the endpoints, then
to be constant. But in that case, we can take any
f
there must be a zero of
[a, b]:
apply Rolle's Theorem to the interval
f 0.
there is necessarily a
c
in
with
Thus, between the two zeroes
a
and
b
of
f,
there is a zero
c
of
f 0.
We can use Rolle's Theorem (and the corollary above) to establish an upper bound on the number of real
roots of a dierentiable function.
◦
When we combine Rolle's Theorem with appropriate use of the Intermediate Value Theorem to show the
existence of real roots, we can often nd the exact number of roots.
•
Example: Show that the function
g(x) = x3 + x − 1
has at least one real root, and then show that
g
cannot
have two real roots.
◦
Notice that
g
conclude that
◦
◦
•
g
g
2
But
◦
Therefore,
g (x) = 3x + 1 is never zero.
g
g(1) = 1.
So by the Intermediate Value Theorem, we
(0, 1).
had another real root: call the two real roots
By the corollary to Rolle's Theorem,
0
while
must have a root somewhere in the interval
Now suppose that
◦
g
0
a
b.
and
then would have to be zero somewhere between
This is an impossibility, meaning that
g
a
and
b.
could not have two real roots.
must have exactly one real root.
Example: Show that the polynomial
◦
g(0) = −1
is continuous, and
p(x) = x7 − 7x + 1
has exactly three real roots.
We use the Intermediate Value Theorem to show the existence of 3 roots, and then Rolle's Theorem (or
more precisely, the corollary given above) to show that there cannot be more than 3 roots.
◦
intervals
◦
p(−2) = −114, p(−1) = 7, p(1) = −5, and p(2) = 115,
(−2, −1), (−1, 1), and (1, 2), meaning it has at least 3 roots.
We compute
•
p
p had 4 or more roots, say a < b < c < d. Then p0 would
(a, b), (b, c), and (c, d) so p0 would have at least 3 roots.
Now suppose that
of the intervals
◦
so
p0 (x) = 7x6 − 7 = 7(x6 − 1) only has two roots, x = 1 and x = −1,
graphing. This is impossible, so p must have exactly 3 roots.
But
has a root on each of the
have to have a zero on each
as can be seen by factoring or
Returning to our discussion of increasing and decreasing functions, we can use Rolle's Theorem to prove a
more general result:
•
Theorem (Mean Value Theorem): If
then there exists some point
c
in
f (x)
(a, b)
is a continuous function on
for which
f (b) − f (a)
f 0 (c) =
.
b−a
8
[a, b]
which is dierentiable on
(a, b),
◦
Equivalently, this means there is a point
f (x)
◦
at
x=c
f 0 (c))
x=c
at
c
in the interval for which the instantaneous rate of change of
is equal to the average rate of change on
Another restatement: there is a point
y = f (x)
◦
(the value
c
in the interval
(a, b)
[a, b]
(the value
f (b)−f (a)
).
b−a
such that the slope of the tangent line to
is equal to the slope of the secant line joining
(a, f (a))
and
(b, f (b)):
Intuitively, from the picture, if we imagine sliding the secant line vertically, we will eventually hit a
point where it is tangent to the graph of
y = f (x).
◦
Proof: The trick is to dene a new function
◦
Also,
◦
So we can apply Rolle's Theorem to
◦
But this means
g(x) to which we can apply
f (b) − f (a)
we use is g(x) = f (x) − x ·
.
b−a
◦ Since f (x) is continuous on [a, b] and dierentiable on (a, b), so is g(x).
Rolle's Theorem. The function
f (b) − f (a)
= [f (b) − f (a)] − [f (b) − f (a)] = 0.
b−a
g(x), which says that there is a c in (a, b) for which g 0 (c) = 0.
g(b) − g(a) = [f (b) − f (a)] − (b − a) ·
f (b) − f (a)
,
b−a
0 = g 0 (c) = f 0 (c) −
so that
f 0 (c) =
f (b) − f (a)
b−a
as we wanted.
3.2.2 Classication of Increasing and Decreasing Behavior
•
Using the Mean Value Theorem, we can make very precise our intuitive ideas about the relationship between
f0
the sign of
•
and the increasing or decreasing behavior of
f.
f (x) is dierentiable on an interval I , and f 0 (x) > 0
f (x) < 0 for all x in I , then f (x) is decreasing on I .
Theorem (Increasing and Decreasing Functions): If
for all
◦
x
in
I,
then
f (x)
is increasing on
Proof: First suppose that
I.
If
f 0 (x) > 0:
on
0
we then want to show that for any
a < b
in
I,
it is true that
f (a) < f (b).
◦
By the Mean Value Theorem applied to
f
on
[a, b],
◦
By assumption
f 0 (x) > 0
I,
so
f 0 (c) > 0.
◦
Therefore
◦
f (b) − f (a)
The argument when
yields some
negative, so
•
By the theorem,
◦
everywhere in
must be positive also, so
0
f (x) < 0
Provided that
critical numbers (the places where
◦
f0
in
b>a
f
0
(a, b)
so
with
b−a
f 0 (c) =
is positive.
and
f
f
0
f 0 (c) < 0
implies that
is decreasing whenever
f0
f (x) = x3 − 3x,
[a, b]
f (b) − f (a)
again
must be
f 0 < 0.
is positive and negative, we may list all of the
is zero or undened) and then plug in a test point in each interval
on that interval.
f0
cannot change sign anywhere else,
has the same sign on each interval.
Example: For
f (b) − f (a)
.
b−a
as desired.
By the Intermediate Value Theorem, we are then guaranteed that
so
•
f 0 > 0,
but now
is continuous, to determine where
to determine the sign of
c
is almost identical: applying the Mean Value Theorem on
is increasing whenever
f0
Also,
f (a) < f (b)
f (b) − f (a)
c in (a, b) with f 0 (c) =
,
b−a
then f (b) < f (a).
f
there exists
determine where
f
is increasing and where
9
f
is decreasing.
f 0 (x) = 3x2 − 3 is continuous, we may use test points. First, we identify the critical numbers
2
setting 3x − 3 = 0 yields 3(x − 1)(x + 1) = 0 so the critical numbers are x = −1 and x = 1.
◦
Since
◦
We therefore have three intervals to consider:
each interval, and determine whether
f0
(−∞, −1), (−1, 1),
of
f:
(1, ∞). We choose a test point in
f 0 necessarily has the same
and
is positive or negative there: then
sign in the entire interval.
◦
(−∞, −1)
On
x = −2:
we can choose the test point
since
f 0 (−2) = 9 > 0,
f0 > 0
we see that
on this
interval.
◦
◦
◦
(−1, 1)
On
we can choose the test point
(1, ∞)
Finally, on
x = 0:
we can choose the test point
since
f 0 (0) = −3 < 0,
x = 2:
since
We can summarize this information with a sign diagram for
we see that
0
f (2) = 9 > 0,
By the discussion above,
f
is therefore increasing on
we see that
here.
0
f >0
here.
0
f : ⊕ | | ⊕.
−1
◦
f0 < 0
(−∞, −1)
and
1
(1, ∞)
and decreasing on
(−1, 1)
.
3.2.3 The First Derivative Test
•
We can also use our analysis of increasing and decreasing behaviors to give a procedure for determining
whether a critical number is a local minimum, local maximum, or neither.
•
Theorem (First Derivative Test): Suppose
c
positive at
(a, c)
◦
for some
f0
If
f0
changes sign from negative to
changes sign from positive to negative at
sign at
c,
then
c
and that
0
f >0
on some other
c, this means that f 0 < 0
interval (c, b) for some c < b.
is negative for numbers slightly smaller than
c,
and
f0
f
has
on some interval
f 0 changes sign from negative to positive at c. This means that f 0 < 0
0
some a < c, and that f > 0 on some other interval (c, b) for some c < b.
Proof: First suppose that
(a, c)
for
By the theorem on increasing and decreasing functions, we can then conclude that
The proof where
c,
interval around
f0
for all
c < y < b.
Therefore,
f (c) ≥ f (z)
for all
on
f (x) < f (c) for all
a < z < b, meaning
changes sign from positive to negative is the same, except with the appropriate
f 0 does not change sign, then f is either increasing
that f has neither a minimum nor a maximum at c.
meaning
or decreasing on an
For reasonable functions, the First Derivative Test says that we can simply read o the type of critical point
from the
•
then
is positive for numbers slightly
inequalities ipped. Finally, if
•
c,
is neither a minimum nor a maximum.
changes sign from negative to positive at
a < x < c, and also that f (c) > f (y)
that c is a local maximum of f .
◦
0
c.
some interval
◦
0
a<c
In other words,
larger than
◦
f has a local minimum at c. If f
0
at c. Finally, if f does not change
When we say f
f.
is a critical number of
, then
a local maximum
◦
c
f0
sign diagram.
Example: For
f (x) = 3x5 − 5x3 ,
or neither), and determine where
◦
◦
nd and classify the critical numbers of
f
is increasing and where
f 0 (x) = 15x4 − 15x2 which is
1)(x + 1) = 0, so there are three critical
numbers:
Next we construct the sign diagram for
f 0:
We have
diagram
Therefore,
◦
Furthermore, at
f
is
0
Setting
x = −1, 0, 1
using test points
f0 = 0
and factoring yields
15x2 (x −
.
x = −2, −1/2, 1/2,
and
2
yields the sign
1
increasing on
x = −1
(−∞, −1)
x = 0, f 0
Finally, at
(1, ∞)
and
we see that
decreasing to increasing, so that
decreasing on
f 0 switches from positive
−1 is a local maximum .
does not change sign, so
x = −1
and
we see that
increasing to decreasing, so that
◦
(local minimum, local maximum,
f : ⊕ | | | ⊕.
◦
At
always dened.
f
is decreasing.
0
−1
◦
f
0
f
1
0
(−1, 0)
and
(0, 1)
to negative, meaning that
f
.
switches from
is neither a minimum nor a maximum .
switches from negative to positive, meaning that
is a local minimum .
10
f
switches from
◦
Here is the graph of
y = f (x),
illustrating what we have found (the critical points are marked, the
increasing portions are in red, and the decreasing portions are in blue):
•
Example: For
p(x) =
1 5 5 3
x − x + 4x + 1,
5
3
nd and classify the critical numbers of
maximum, or neither), and determine where
f
is increasing and where
f
f
(local minimum, local
is decreasing.
p0 (x) = x4 − 5x2 + 4 = (x2 − 1)(x2 − 4) = (x − 1)(x + 1)(x − 2)(x + 2).
◦
We have
◦
Thus, the critical numbers are
◦
Plugging in test points (e.g.,
x = −2, −1, 1, 2
, since
x = −3, −1.5, 0, 1.5, 3)
p0
is dened everywhere.
yields the sign diagram
⊕ | | ⊕ | |⊕
−2
◦
Thus,
p
◦
Also,
x = −1
and
◦
is
increasing on
x = −2
(−∞, −2), (−1, 1),
and 2 are local minima
and 1 are local maxima
Here is the graph of
y = p(x),
since
since
and
p
p
(2, ∞)
, and
decreasing on
−1
(−2, −1)
1
and
for
p0 .
2
(1, 2)
.
switches from decreasing to increasing at both locations,
switches from increasing to decreasing at both locations.
illustrating what we have found (the critical points are marked, the
increasing portions are in red, and the decreasing portions are in blue):
3.3
•
Concavity, Graphing With Calculus
f 0 of a function f to analyze its behavior: we now
00
turn our attention to studying what the second derivative f can tell us. Then we will explain how to use all
of this information to draw an accurate graph of y = f (x).
We have already discussed how to use the rst derivative
3.3.1 Concavity, Inection Points, and the Second Derivative
•
By denition,
f 00
represents the rate of change of the derivative
11
f 0.
◦
f 00 > 0
Thus, the statement that
on an interval is equivalent to saying that the function
f0
is increasing
on that interval.
◦
Because
f0
that the slopes of the tangent lines to the graph of
◦
y = f (x),
represents the slope of the tangent line to the graph of
y = f (x)
derivative is
f (x) = 2:
f 00 > 0
says
are increasing.
We can see this quite clearly from the graph of the function
00
the statement
y = f (x)
for
f (x) = x2 ,
whose second
in this graph, the tangent slopes clearly increase as we move from left to right
along the plot, producing an overall upward-opening shape that rather closely matches the actual shape
of the graph of
◦
y = x2 :
By the same logic, if
f 00 < 0,
then the tangent slopes to the graph of
y = f (x)
should be decreasing,
suggesting that the graph should have a downward-opening shape.
◦
This intuition is borne out by the graph (above) of
y = f (x)
for
f (x) = −x2 ,
whose second derivative is
00
f (x) = −2.
•
Let us give a more precise denition of this interpretation of the behavior of
f 0 (namely, whether it is increasing
or decreasing):
•
Denition: If
on
•
I,
f
is a dierentiable function dened on an interval
and we say
f
is concave down on
I
if
f
0
is decreasing on
I,
we
f
is concave up on
I
if
f0
is increasing
I.
As we already worked out intuitively above, the concavity of a function is related to the sign of its second
derivative:
•
f (x) is twice dierentiable on an
f (x) < 0 for all x in I , then f (x) is
Theorem (Concavity of Functions): If
I,
then
◦
f (x)
is concave up on
Proof: Let
g = f 0:
I.
If
0
interval
I,
f 00 (x) > 0
on I .
and
concave down
then by our earlier theorem about increasing functions, if
g0 > 0
on
for all
I
then
x
in
g
is
increasing.
◦
◦
Rephrasing in terms of
0
If
g <0
f
is concave down.
then
g
•
The locations where
•
Denition: If
◦
f
f:
if
f 00 > 0
on
I,
then
f0
is increasing, or, equivalently,
is decreasing, which is equivalent to saying that if
f
00
f <0
on
I,
f
is concave up.
then
f0
is decreasing, so
changes concavity have a special name:
changes concavity at a point, that point is called a point of inection (or inection point).
Points of inection are sometimes called turning points: when drawing the graph, you will nd that you
must turn your hand when the curve passes through a point of inection.
◦
If
f0
exists everywhere, then the points of inection are precisely the places where
12
f 00
changes sign.
•
Let us give a few examples of concavity and points of inection:
◦
Example: Because
up on
◦
(0, ∞),
f 00
f (x) = x4 − 6x2 has f 00 (x) = 12(x2 − 1), we see that f is concave up on (−∞, −1) ∪
(−1, 1). Furthermore, because f 00 changes sign at x = −1 and x = 1, it has
inection at (−1, −5) and (1, −5).
is continuous, then to nd points of inection and study concavity, we can make a sign diagram for
Explicitly, we rst mark o all places where
no other places where
◦
f 00
00
>0
are where
places where
f 00
f
is zero or undened. By the continuity of
is concave up, the intervals where
p0 (x) = 4x3 − 6x2
the inection points of
00
f
<0
We have
◦
Thus, the potential points of inection are
◦
Plugging in test points (e.g.,
are where
and
f,
◦
Also, since
for
◦
is
concave up on
p00
there.
f
The intervals where
is concave down, and the
x = 0, 1,
x = −1, 0.5, 2)
(−∞, 0),
switches sign at
f
the intervals where
since
p00
is concave up, and the
is dened everywhere.
yields the sign diagram
⊕| | ⊕
0
p
f 00
p00 (x) = 12x2 − 12x = 12x(x − 1).
◦
Thus,
there are
changes sign are the inection points.
p(x) = x4 − 2x3 , nd
where f is concave down.
◦
f 00 ,
could change sign.
Example: For
intervals
f 00
.
We then plug in a test point in each interval to determine the sign of
f
•
f
00
0
f
using test points in exactly the same way that we did for
◦
(−∞, 0) and concave
and concave down on
points of
If
is concave down on
Example: Because
(1, ∞)
•
and
f (x) = x3 −3x has f 00 (x) = 6x, we see that f
has a point of inection at (0, 0).
and
(1, ∞)
, and
concave down on
x = 0 and at x = 1, we see that (0, 0)
for
p00 .
1
(0, 1)
and
.
(1, −1)
are points of inection
p.
Here is the graph of
y = p(x),
illustrating what we have found (the inection points are marked, the
concave-up portions are in purple, and the concave-down portions are in green):
13
•
2
f (x) = e−x /2 , nd
where f is concave down.
Example: For
intervals
◦
We have
◦
00
Since
when
◦
f 0 (x) = −xe−x
f is always
x = ±1.
2
/2
the inection points of
, and
2
f 00 (x) = −e−x
/2
f,
the intervals where
+ (−x)(−x)e−x
2
/2
= (x2 − 1)e−x
x = −2, 0, 2)
yields the sign diagram
⊕ | |⊕
−1
Thus,
◦
Also, since
for
◦
f
is
is concave up, and the
2
/2
dened, we see that the only points of inection could occur when
Plugging in test points (e.g.,
◦
f
concave up on
(−∞, −1),
and
(1, ∞)
, and
for
.
f 00 (x) = 0;
namely,
f 00 .
1
concave down on
(−1, 1)
f 00 switches sign at x = −1 and at x = 1, we see that (−1, e−1/2 )
and
.
(1, e−1/2 )
are points of inection
f.
Here is the graph of
y = f (x),
illustrating what we have found (the inection points are marked, the
concave-up portions are in purple, and the concave-down portions are in green):
◦
Remark: The curve analyzed above is (up to a scaling factor) the famous Gaussian normal distribution,
commonly called the bell curve. It shows up very frequently in statistics.
3.3.2 Geometric Properties of Concavity, the Second Derivative Test
•
There are some other useful geometric interpretations of concavity. Here are two of them:
•
Theorem (Concavity and Secant Lines): A function is concave up on an interval if it lies below its secant lines:
in other words, if the portion of the graph between
(b, f (b)).
x=a
and
x=b
lies below the line joining
◦
Here are illustrations of this property:
◦
Proof: Suppose rst that
◦
Observe that the given statement is equivalent to saying that for any
f
and
is concave up.
b−t
t−a
f (a) +
f (b). (The right-hand
b−a
b−a
line joining (a, f (a)) and (b, f (b)).)
◦
(a, f (a))
A function is concave down if it lies above its secant lines.
term is the
y -coordinate
This inequality can be rearranged into the equivalent form
the Mean Value Theorem: the left term is equal to
14
f 0 (c)
for
t
with
a < t < b,
of the point with
one has
f (t) <
x-coordinate t
on the
f (t) − f (a)
f (b) − f (t)
<
, which holds by
t−a
b−t
some c in (a, t), and the right term is equal
to
f 0 (d)
for some
d
in
(t, b):
then because
f 00 (x) > 0,
we know that
f0
is increasing, so
f 0 (c) < f 0 (d),
as
desired.
◦
•
For concave-down functions the argument is similar, except with the appropriate inequalities ipped.
Theorem (Concavity and Tangent Lines): A function is concave up on an interval if it lies above its tangent
lines (except for the points of tangency), and a function is concave down on an interval if it lies below its
tangent lines.
◦
◦
Here are illustrations of this property:
f is concave up. Observe that the given statement is equivalent to the statement
f (t) > f (a) + f 0 (a) · (t − a) for any t 6= a. (The right-hand term is the y -coordinate of the point
with x-coordinate t on the tangent line to y = f (x) at x = a.)
f (t) − f (a)
◦ For t < a, this inequality is equivalent to
< f 0 (a), which holds by the Mean Value Theorem:
t−a
0
0
0
0
the left term is equal to f (c) for some c in (t, a), and f (c) < f (a) by the assumption that f is increasing.
f (t) − f (a)
◦ Similarly, if t > a, then the inequality is equivalent to f 0 (a) <
, which again holds by the
t−a
0
0
0
Mean Value Theorem: the left term is equal to f (d) for some d in (a, t), and f (a) < f (d) by the
0
assumption that f is increasing.
◦ For concave-down functions the argument is similar, except with the appropriate inequalities ipped.
Proof: Suppose rst that
that
•
As a nal remark, there is a way to use the second derivative to determine whether a critical number is a
local minimum or local maximum.
◦
The method we previously discussed, of determining whether
f
changes sign at a critical number, will
always work at least as well as this test does. (This test does have the advantage of occasionally requiring
slightly less computation.) We include this method only for completeness.
•
Theorem (Second Derivative Test): If
f 0 (c) = 0),
◦
then
c
Remark: If
f
is a local maximum if
f 00 (c) = 0,
c a critical
f 00 (c) > 0.
is a twice-dierentiable function with
f 00 (c) < 0
and
c
is a local minimum if
number (so that
the test yields no information. There could be a local minimum, local maximum,
or neither.
◦
Intuitively, if
f 00 > 0
then
f
is concave up, so any critical number must be a local minimum because the
graph must open upwards. Inversely, if
f 00 < 0
then
f
is concave down, so any critical number must be
a local maximum because the graph must open downwards.
◦
◦
Proof: First suppose
and
f 0 (c) = 0:
then
f 00
is positive on some open interval containing
By the theorem on concavity and tangent lines, the graph of
line at
◦
f 00 (c) > 0
x=c
on an open interval containing
y = f (x) lies
c.
above the graph of its tangent
c.
f (x) ≥ f (c) on an open interval
c, which means that c is a local minimum.
◦ If f 00 (c) < 0 and f 0 (c) = 0, then in a similar way we conclude that f lies below the graph of its horizontal
tangent line at c, meaning f (x) ≤ f (c) and so c is a local maximum.
But the tangent line is horizontal, so we immediately conclude that
containing
•
Example: Use the Second Derivative Test to identify the critical numbers of
f (x) = x3 − 12x
as local minima
or local maxima.
◦
We have
◦
00
Since
so
2
f 0 (x) = 3x2 − 12 = 3(x − 2)(x + 2),
so the critical numbers are
00
f (x) = 6x, we see f (−2) = −12 < 0, meaning that −2
is a local maximum .
15
x = −2
and
x = 2.
is a local maximum , and
f 00 (2) = 12 > 0,
3.3.3 Graphing Functions Using Calculus
•
By analyzing the rst and second derivatives of a function
f:
f,
we can obtain a great deal of information about
we can determine the locations of critical points and inection points, classify relative minima and maxima,
and determine where the function is increasing, decreasing, concave up, and concave down.
•
Using this information, we can plot graphs of functions more precisely (and usually, more easily) than the
standard procedure of plugging many points into the function and joining them with a curve.
◦
Generally speaking, the most interesting features of the graph of
y = f (x) are places near critical points or
f 0 and f 00 and determining
inection points. We can identify these locations by studying the derivatives
where the values change sign.
◦
Once we have identied the approximate locations of interesting features of the graph, we can ll in
the portion of the graph between those points: we will know fairly accurately how the function behaves
between its critical points and inection points (i.e., whether it is increasing or decreasing, and whether
it is concave up or concave down).
•
If the graph of
y = f (x)
approaches a line (in some manner), we call that line an asymptote of the graph.
Specically:
◦
If
lim f (x) = ∞
or
x→a+
y = f (x).
−∞,
or
lim f (x) = ∞
ln(x)
x = a
is called a vertical asymptote of
x = 0.
at
lim [f (x) − (Ax + B)] = 0 (respectively, lim [f (x)
x→∞
x→−∞
called a slanted asymptote (or horizontal asymptote if
First, analyze
− (Ax + B)] = 0),
A = 0)
1
x
at
x = 0,
y = Ax + B
then the line
to the graph of
but they
y = f (x)
as
is
x → ∞
x → −∞).
To identify all of the important features of a graph
◦
the line
If
(respectively, as
•
−∞,
Typically, vertical asymptotes occur because of division by zero, e.g.,
can also occur for other functions like
◦
or
x→a−
f0
y = f (x),
we can follow these steps:
to identify any local minima or maxima, and the intervals where
f
is increasing or
decreasing.
∗
∗
Find all critical points by determining when
determine whether
When
f 0 > 0, f
f
0
−∞
f 0 (x)
is undened.
Furthermore, at each critical point, if
f0
f 00
f
f 0 < 0, f
0
+∞),
f0
plug in a test value to
to
is decreasing.
changes sign from positive to negative then
changes sign from negative to positive then
not change sign, then
Next, analyze
and the interval out to
is positive or negative on that interval.
is increasing, and when
maximum, and if
◦
and when
Mark all the critical points on a number line, and then in each interval between two critical points
(as well as the innite interval out to
∗
∗
f 0 (x) = 0,
f
f
f
has a local
has a local minimum. (If
f0
does
has neither a minimum nor a maximum.)
f
to identify any points of inection, and the intervals where
is concave up or concave
down.
∗
∗
Find
f 00 (x)
and set it equal to zero, to nd all potential points of inection.
Mark all potential points of inection on a number line (along with all points where
and then plug in test points to
∗
∗
◦
When
When
f
00
to determine the sign of
f 0 > 0, f is concave up, and when f 0 < 0, f
f 00 changes sign, f has an inection point.
Then, analyze the behavior of
f (x)
for large
x,
f
00
f 00
is concave down.
and look for any asymptotes.
∗
Find
lim f (x) and lim f (x). If the limit as x → −∞ or x → ∞ has a nite value
x→−∞
x→∞
a horizontal asymptote y = L as x → −∞ or x → ∞ (as appropriate).
∗
To nd vertical asymptotes, search for values of
equal to
∞
or
−∞.
16
is undened),
on each interval.
c
such that
lim f (x)
x→c−
or
lim f (x)
x→c+
L,
then
f
has
(or both) are
∗
A,
number B ,
x → −∞.
number
◦
f (x)
.
x
lim [f (x) − Ax].
To nd slanted asymptotes, rst compute
then compute the limit
then
y = Ax + B
lim
If this rst limit exists and is a nite nonzero
x→∞
If this second limit also exists and is a nite
x→∞
is a slanted asymptote as x → ∞. Repeat the process with limits as
Finally, assemble all of the information (increasing/decreasing behavior, minima and maxima, concavity,
points of inection, behavior for large
∗
x,
asymptotes) to draw the graph.
Begin by computing the coordinates of all of the critical and inection points and plotting them
accurately in the plane.
∗
Next, on each of the
x-intervals between consecutive pairs of the plotted points, identify the behavior
of the function on that interval (increasing and concave up, increasing and concave down, decreasing
and concave up, or decreasing and concave down) and draw an appropriately-shaped curve joining
the two endpoints of the interval.
∗
Finally, use the information about the behavior as
for the remaining values of
•
x → −∞
x→∞
and
to plot the behavior of
f (x) = x3 + 3x2 + 1. Find the critical points, local minima and maxima,
and the intervals where f is increasing, decreasing, concave up, or concave down. Then
y = f (x).
Example: Let
◦
First, we analyze
f
x.
points of inection,
sketch the graph of
f 0 (x) = 3x2 + 6x = 3x(x + 2).
f 0 (x) = 0
∗
Setting
∗
Next, we draw the number line, mark o the two critical points, and then plug in test points (for
example, we could use
⊕ | | ⊕.
3x(x + 2) = 0,
yields
so we see that
x = −3, x = −1,
and
x = −2, 0
x = 1)
are critical points .
f 0 sign diagram looks like
f (x) = 3x(x + 2) to make the sign
to see that the
0
Alternatively, we could have used the factored form
−2
0
diagram.
∗
From this, we conclude that
∗
Also, since
f0
◦
f0
is
increasing on
Setting
(−∞, −2)
changes from positive to negative at
changes from negative to positive at
Next, we analyze
∗
∗
f
and
(0, ∞)
x = −2, (−2, 5) is
x = 0, (0, 1) is
and
decreasing on
(−2, 0)
.
a local maximum , and since
a local minimum .
f 00 (x) = 6x + 6.
f 00 (x) = 0
yields
x = −1.
We draw the number line, mark o
diagram looks like
x = −1,
and then plug in two test points to see that the
f 00
sign
| ⊕.
−1
∗
From this, we conclude that
since
◦
Since
f
f 00
changes sign at
f
is
x = −1
concave up on
we see that
(−1, 3)
and
f
concave down on
(−∞, −1)
, and
is a point of inection .
is a polynomial of odd degree with positive leading coecient,
There are also no asymptotes, again because
◦
(−1, ∞)
lim f (x) = −∞ and lim f (x) = ∞.
x→−∞
x→∞
is a polynomial.
Now we can assemble the graph.
∗
From
x = −∞, f
is concave down and increasing from
−∞,
until it hits a local max at
(−2, 5)
and
begins decreasing.
∗
The function has a point of inection at
until it hits a local min at
∗
∗
Then
f
(−1, 3),
switching to concave up, and continues decreasing
(0, 1).
increases and is concave up out to
x = +∞.
Here is the resulting graph (with the local extrema and point of inection marked):
17
•
h(x) = 2x6 − 3x4 . Find the critical numbers, local minima and maxima, points of inection,
intervals where f is increasing, decreasing, concave up, or concave down, the absolute minimum and maximum
of h on the interval−2 ≤ x ≤ 2, and nally, sketch the graph of y = h(x).
Example: Let
◦
h0 (x) = 12x5 − 12x3 = 12x3 (x2 − 1) = 12x3 (x + 1)(x − 1).
First, we analyze
h0 (x) = 0
12x3 (x + 1)(x − 1) = 0,
x = −1, 0, 1
∗
Setting
∗
Next, we draw the number line, mark o the three critical points, and then plug in test points (for
yields
example, we could use
like
| ⊕ | | ⊕.
−1
0
∗
Thus,
∗
Also, since
f
is
so we see that
x = −2, x = −1/2, x = 1/2,
and
x = 2)
are critical numbers .
f 0 sign diagram looks
h0 (x) = 12x3 (x + 1)(x − 1).
to see that the
Alternatively, we could have used the factorization
1
increasing on
f0
(−1, 0)
and
(1, ∞)
and
changes from positive to negative at
decreasing on
x = 0, (0, 0)
(−∞, −1)
and
(0, 1)
.
is a local maximum .
f 0 changes from negative to positive at x = −1 and x = 1, (−1, −1) and (1, −1) are local minima
p
p
◦ Next, we analyze h00 (x) = 12(5x4 − 3x2 ) = 12x2 (5x2 − 3) = 60x2 (x − 3/5)(x + 3/5).
p
∗ Setting h00 (x) = 0 yields x = 0, ± 3/5.
∗ We draw the number line, mark o the three points, and then plug in test points to see that the f 00
sign diagram looks like ⊕ |
√ | √| ⊕. (Again, we could also have used the factorization to help.)
∗
Since
3
5
−
∗
◦
0
3
5
(−∞, −1) and (1, ∞) and concave down on (−1, −0) and (0, 1) ,
p
p
00
since f changes sign at x = ±
3/5 but not at x = 0, we see that (± 3/5, −81/125) are points
Thus,
Since
h
f
is
concave up on
is a polynomial of even degree with positive leading coecient,
and
of inection .
lim h(x) = ∞ = lim h(x),
x→−∞
x→∞
and there are no asymptotes.
◦
Now we may assemble the graph.
∗
From
x = −∞, h
is concave up and decreasing from
∞,
until it hits a local min at
(−1, −1)
and
begins increasing.
∗
The function has a point of inection at
increasing until it hits a local max at
(−
q
3
81
5 , − 125 ), switching to concave down, and continues
(0, 0), where the function attens out (though it doesn't change
concavity).
∗
Next
h
begins decreasing, but remains concave down to
h
hits a local min at
(
q
3
81
5 , − 125 ), where it switches to concave
up.
∗
∗
Then
(1, −1),
.
and then nally increases (remaining concave up) out to
Here is the resulting graph (with the local extrema and point of inection marked):
18
x = ∞.
•
Example: Let
f (x) = x+
1
.
x
Find the critical numbers, local minima and maxima, points of inection, vertical
and slanted asymptotes, and the intervals where
Then sketch the graph of
◦
First, we analyze
f
is increasing, decreasing, concave up, or concave down.
f0
is undened at
y = f (x).
1
.
x2
f 0 (x) = 1 −
f 0 (x) = 0
Note that
x2 = 1
(as is
f
itself ).
x = −1, 1
∗
Setting
∗
Next, we draw the number line, mark o the two critical numbers and the value
yields
x = 1, −1,
x=0
so that
so we see that
undened, and then plug in test points to see that the
f
0
are critical numbers .
sign diagram looks like
x = 0 where f 0
| ⊕ | ⊕ | .
−1
∗
Thus,
∗
Also, since
f
◦
0
f
is
increasing on
f0
and
(0, 1)
Note that
f 00
and
decreasing on
changes from negative to positive at
changes from positive to negative at
Next, we analyze
∗
∗
(−1, 0)
f 00 (x) =
(−∞, −1)
x = −1, (−1, −2) is
x = 1, (1, 2) is
and
(1, ∞)
0
is
1
.
a local minimum , and since
a local minimum .
2
.
x3
is never zero, so there are no points of inection.
To analyze concavity, we draw the number line, mark o
plug in two test points to see that the
f
00
x=0
sign diagram looks like
(where
f 00
is undened), and then
| ⊕.
0
∗
◦
From this, we conclude that
We also easily nd
is
lim f (x) = −∞
x→−∞
∗
There is a vertical asymptote at
∗
Also, since
since
◦
f
lim [f (x) − x] = 0
x→∞
lim [f (x) − x] = 0
x→−∞
concave up on
and
(0, ∞)
and
concave down on
(−∞, 0)
.
lim f (x) = ∞.
x→∞
x = 0, since we can compute lim f (x) = −∞ and lim f (x) = +∞.
x→0−
we see that
we see that
y =x
y=x
x→0+
is a slanted asymptote as
is also a slanted asymptote as
x → ∞,
and likewise
x → −∞.
Now we can assemble the graph.
∗
(−1, −2) and then
−∞ as x approaches 0 from below.
∗ There is a vertical asymptote at x = 0, and on the positive side of 0, f is concave up and decreasing
(from ∞) until it hits a local min at (1, 2), and then increases to ∞ as x → ∞.
∗ Furthermore, the function approaches the line y = x as x → −∞ and as x → ∞.
∗ Here is the resulting graph (with the local extrema and asymptotes marked):
From
x = −∞, f
is concave down and increasing until it hits a local max at
decreases back down towards
19
3.4
•
L'Hôpital's Rule
By employing derivatives, we can evaluate certain kinds of indeterminate limits whose computations are not
susceptible to the basic limit techniques. The most important theorem for evaluating such limits is L'Hôpital's
rule.
•
L'Hôpital's Rule: If
diverge to
◦
◦
∞
or
−∞
f (x)
as
and
g(x)
x → a,
are dierentiable functions and either
then
f 0 (x)
f (x)
= lim 0
,
lim
x→a g (x)
x→a g(x)
The rule also holds for one-sided limits, or if
Proof (of special case where
a
f (a) = g(a) = 0
is
+∞
and
f (a) = g(a) = 0,
or both
f
and
g
assuming the second limit exists.
or
−∞.
0
g (a) 6= 0):
By the division rule for limits, and the
limit denition of the derivative, we have
f (x) − f (a)
f (x) − f (a)
limx→a
f (x)
f (x) − f (a)
f 0 (a)
x−a
x−a
=
= 0
lim
= lim
= lim .
x→a g(x)
x→a g(x) − g(a)
x→a
g(x) − g(a)
g(x) − g(a)
g (a)
limx→a
x−a
x−a
◦
The other cases of the rule require a more intricate and careful proof. There are several approaches; one
is to use a result known as Cauchy's Mean Value Theorem.
•
Important Warning:
L'Hôpital's rule is not a magic formula for evaluating limits, and there are many
indeterminate limits for which L'Hôpital's rule is not suitable. In many such cases, there are other techniques
available that provide a far simpler and easier solution than L'Hôpital's rule, and in other cases L'Hôpital's
rule can fail to say anything at all.
•
Example
0
:
0
Evaluate
◦
At
◦
We obtain
x = 0,
both
lim
x→0
tan(2x)
tan(2x)
.
x
and
x
take the value zero, so we can apply L'Hôpital's rule.
tan(2x)
2 sec2 (x)
= lim
= lim 2 sec2 (x) = 2 sec2 (0) = 2
x→0
x→0
x→0
x
1
lim
.
•
In some cases, we may need to apply L'Hôpital's rule several times before obtaining a limit we can evaluate.
•
Example
0
:
0
Evaluate
◦
At
◦
We obtain
◦
We get
x = 0,
lim
x→0
sin(x) − x
.
x3
both the numerator and denominator are zero, so we can apply L'Hôpital's rule.
sin(x) − x
cos(x) − 1
0
= lim
. The new limit is a
limit so we use L'Hôpital's rule again.
x→0
x3
3x2
0
cos(x) − 1
− sin(x)
0
lim
= lim
. This is still a
limit, so we use L'Hôpital's rule a third time.
x→0
x→0
3x2
6x
0
lim
x→0
20
◦
•
We nally get
Example
◦
As
∞
∞
x
− sin(x)
− cos(x)
1
= lim
=− .
x→0
6x
6
6
lim
x→0
Thus, the original limit
1
sin(x) − x
= −
3
x
6
x2
.
x→∞ ex
tends to innity,
ex
and
x2
both tend to
+∞,
so we can apply L'Hôpital's rule.
2x
x
= lim x . The
x→∞ e
ex
2
2x
= lim x = 0.
lim
x→∞ e
x→∞ ex
∞
∞
◦
We obtain
◦
This yields
◦
Remark: By repeated application of L'Hôpital's rule, we can see in a similar way that
lim
of
Thus, the original limit
limit so we use L'Hôpital's rule again.
x2
= 0
x→∞ ex
lim
.
xn
= 0 for
x→∞ ex
x
n
this says that e grows more rapidly than x for any value
n (even non-integral values of n):
lim
n.
The two limit forms
◦
new limit is still an
x→∞
any value of
.
lim
: Evaluate
2
•
lim
x→0
0
0
and
∞
∞
appearing in the statement of L'Hôpital's rule are called indeterminate limit forms.
Some other indeterminate forms are
0 · ∞, ∞ − ∞, 00 , ∞0 ,
and
1∞ .
When these other forms occur, we
will need to rearrange the given limit before L'Hôpital's rule can be applied.
•
Example
◦
As
(0 · ∞):
x
Evaluate
tends to innity,
h
πi
.
lim x · tan−1 (x) −
x→∞
2
tan−1 (x) −
π
2
tends to zero from below, so this is a
tan−1 (x) −
◦
We can rearrange the limit as
lim
1/x
x→∞
π
2,
tan−1 (x) −
◦
◦
Applying L'Hôpital's rule gives
lim
1/x
x→∞
and now it is a
0
0
0·∞
form.
form.
π
2
2
2 = lim 1/(1 + x ) = lim −x .
x→∞ −1/(x2 )
x→∞ 1 + x2
−x2
−2x
= lim
= lim (−1) = −1. (Alternatively, we
2
x→∞ 1 + x
x→∞ 2x
x→∞
2
could have used the polynomial trick and divided top and bottom by x . Using L'Hôpital's rule repeatedly
Applying L'Hôpital's rule again gives
lim
will give the same result as that trick.)
◦
•
Thus we see that
Example
◦
As
(∞ − ∞):
x
tends to
h
πi
lim x · tan−1 (x) −
= −1
x→∞
2
Evaluate
π
2
lim
x→(π/2)−
[sec(x) − tan(x)].
from below, both
1
◦ Since sec(x) =
cos(x)
0
now a
form.
0
and
sec(x)
and
sin(x)
tan(x) =
,
cos(x)
•
.
tan(x)
blow up to
+∞,
so this is an
we can rewrite the limit as
lim
x→(π/2)−
Applying L'Hôpital's rule gives
◦
So we see that
◦
Remark: In fact, the two-sided limit exists and is zero, by the same calculation.
Example
◦
(00 ):
lim
x→(π/2)−
Evaluate
x→(π/2)−
form.
1 − sin(x)
cos(x)
1 − sin(x)
− cos(x)
0
= lim −
= = 0.
cos(x)
− sin(x)
1
x→(π/2)
◦
lim
∞−∞
[sec(x) − tan(x)] = 0
.
lim xx .
x→0+
We cannot apply L'Hôpital's rule directly, because the limit is of the indeterminate form
21
00 .
, and it is
◦
However, if we take the natural log, we get
lim+ xx = lim+ [ln(xx )] = lim+ [x ln(x)],
ln
x→0
x→0
logarithm is continuous and thus we can move it through limits.
◦
As
x
as written because it is not of the form
◦
If we rewrite the limit as
of the form
•
∞
.
∞
lim
x→0+
0
0
◦
So we get
◦
Exponentiating gives the original limit as
◦
we have a limit of the form
0 · ∞.
ln(x)
1/x
= lim
= lim (−x) = 0.
+
1/x
x→0 −(1/x2 )
x→0+
lim x ln(x) = ln lim xx = 0.
Applying the rule gives
(1∞ ):
We cannot apply L'Hôpital's rule to the limit
ln(x)
, now we can apply L'Hôpital's rule, because now we have something
1/x
◦
Example
ln(x) → −∞.
∞
or
: instead,
∞
tends to 0 from the positive direction,
because the
x→0
lim
x→0+
x→0+
Evaluate
x→0+
lim
t→∞
1+
x t
,
t
where
lim xx = 1
x→0+
x
.
is an arbitrary real number.
This limit does not seem like one to which we can apply L'Hôpital's rule: it is of the indeterminate form
1∞ .
◦
However, if we take the natural logarithm, we get
h
x t
x t
x i
ln lim 1 +
= lim ln 1 +
= lim t ln 1 +
,
t→∞
t→∞
t→∞
t
t
t
because the logarithm is continuous and thus we can move it through limits.
x
→ 0, so we have a limit of the form 0 · ∞. In order to apply L'Hôpital's
t → ∞, ln 1 +
t
x
ln 1 +
t .
rewrite the limit as lim
t→∞
1/t
x
x
ln 1 +
(−x/t2 )/(1 + )
x
t = lim
t = lim
◦ Applying the rule gives lim
x = x.
2
t→∞
t→∞
t→∞
1/t
−1/t
1+
t
t
x
◦ Finally, exponentiating gives the original limit as lim 1 +
= ex .
t→∞
t
◦
As
◦
Note: This limit is often taken as the denition of the (natural) exponential function
x = 1)
•
the denition of the number
or (by setting
e.
Example (L'Hôpital's rule does not work): Find
◦
ex ,
rule, we
eln(x)
.
x→∞
x
lim
If we plug in, we see that this limit is of the form
∞
.
∞
e
eln(x)
◦ If we try to use L'Hôpital's rule, we obtain lim
= lim
x→∞
x→∞
x
1
ln(x)
x = lim e
:
x→∞
1
x
ln(x)
·
this is exactly the
same limit we started with! (Applying the rule more times will, of course, not change this.)
◦
•
The correct way to evaluate this limit is to observe that
Example (L'Hôpital's rule makes a limit worse): Find
eln(x) = x,
and so the limit is just
lim
x→∞
x
= 1
x
sin10 (x)
.
x→0
x10
lim
0
.
0
sin10 (x)
10 sin9 (x) · cos(x)
sin9 (x) · cos(x)
lim
=
lim
=
lim
.
x→0
x→0
x→0
x10
10x9
x9
◦
If we plug in, we see that this limit is of the form
◦
Let us use L'Hôpital's rule: we obtain
22
.
◦
When we attempt to plug in here, we see the limit is still of the form
8
9
10
2
0
,
0
so we try applying the rule
9 sin (x) · cos (x) − sin x
0
sin (x) · cos(x)
= lim
, which is still of the form
.
x→0
x9
9x8
0
72 sin7 (x) cos3 (x) − 28 sin9 (x) cos(x)
9 sin8 (x) · cos2 (x) − sin10 x
=
lim
,
◦ Applying the rule again yields lim
x→0
x→0
9x8
72x7
0
which is still of the form
.
0
504 sin6 (x) cos4 (x) − 468 sin8 (x) cos2 (x) + 28 sin10 (x)
◦ If we try once more, we will end up with lim
, after
x→0
504x6
again: we get
lim
x→0
rather more arithmetic.
◦
It appears that we are making very little progress at the expense of a great deal of algebra: in fact,
it turns out that we would need to apply L'Hôpital's rule another 6 times before we could plug in to
evaluate the limit (and the odds of making some kind of algebra or arithmetic mistake somewhere before
that point is extremely high).
◦
A much simpler way to evaluate this limit is to use the limit laws to reduce this limit to a simpler
one: explicitly, we can write
sin10 (x)
=
lim
x→0
x10
sin(x)
lim
x→0
x
10
= 110 = 1
, where we used the evaluation
sin(x)
lim
= 1 (which we computed using geometry back when we found the derivative of sine, or could
x→0
x
alternatively evaluate with a single application of L'Hôpital's rule).
3.5
•
Applied Optimization (Functions of One Variable)
In many applications (particularly ones with an economic or physical avor), we are interested in optimizing
some quantity: maximizing prot, minimizing waste, etc.
•
By our analysis of minimum and maximum values, in order to maximize or minimize a function of one variable,
we need only nd the function's critical points, and then search among the resulting function values to nd
the maximum or minimum we are interested in.
◦
Frequently there will be physical constraints on the variables involved; e.g., a length cannot be negative.
With such constraints we also need to consider the boundary behavior, and the behavior as the variable
tends to
◦
−∞
∞
or
(as appropriate).
If there are several variables involved, we will need to choose one, and use the given information to
eliminate all variables but one, so that our function of interest is of one variable only, because our
current methods
•
1 cannot accommodate optimization problems involving more than one variable.
Most of the work in solving an applied optimization problem occurs in translating the given information into
a calculus problem. Here is a general procedure for solving such problems:
◦
Step 1: If not already given, choose variable names and translate the given information into mathematical
statements. Draw a picture, if necessary.
•
◦
Step 2: Identify the target function (the one to be optimized) and any constraints.
◦
Step 3: Solve the constraints to write the target function as a function of a single variable.
◦
Step 4: Dierentiate the target function with respect to its variable, and nd the critical points.
◦
Step 5: From the critical points, identify the minimum or maximum of the target function.
r dm and height h dm. If the
π liters, what is the minimum possible surface area for the jug? (Note: 1 dm = 10 cm,
Example: A cylindrical jug, with a bottom and sides but no top, has radius
volume of the jug is to be
and
◦
1 L = 1 dm3 .)
The volume of the jug is
has area
1 Using
2
πr dm
2
πr2 h dm3 ,
π(r2 + 2rh) dm2 ,
2
area 2πrh dm .
and the surface area is
and the lateral surface of the jug has
because the base of the jug
multivariable calculus, one can solve such optimization problems involving multiple variables directly, without having to
reduce to having a function of only one variable.
23
◦
Our target function is the surface area
Since
◦
r
and
2
π r +
r
2
2−
h in terms of r (since it is easier) to get h =
Then
1
SA = π r2 + 2r · 2 =
r
.
SA(r)
2
= 0,
r2
We can see that
with respect to
so that
r=1
d(SA)
2
r:
= π· 2− 2 .
dr
r
Setting this expression equal to zero
r = 1.
is a minimum for the surface area (since if
00
SA (1)
be large; alternatively, we can check that
minimum surface area is
•
1
.
r2
We dierentiate
yields
◦
are lengths, they are positive.
We solve the constraint for
◦
h
SA = π(r2 + 2rh), and we also have the constraint π = V = πr2 h.
2
SA(1) = 3π dm
Example: The three vertices of a triangle are
r
is large or small then
r2 +
r = 1,
is positive). So the minimum occurs at
2
r
will
and the
.
(1, 1), (5, 5),
and
(x, 7).
Find the value of
x
which minimizes
the triangle's perimeter.
◦
The perimeter is the sum of the three side lengths of the triangle. By the distance formula, the perimeter
is
P (x) =
p
(5 − 1)2 + (5 − 1)2 +
p
p
(x − 1)2 + (7 − 1)2 + (x − 5)2 + (7 − 5)2 ,
we wish to minimize.
Expanding the squares gives the slightly simpler expression
◦
We then compute
◦
Cancelling a factor of 2 from each fraction and rearranging gives
P (x) =
2x − 10
◦
x−1
x−5
p
= −p
,
2
(x −1) + 36
(x − 5)2 + 4
(x − 1)2 (x − 5)2 + 4 = (x − 5)2 (x − 1)2 + 36 .
(x − 1)2 (x − 5)2 term from both
2(x − 1) = ±6(x − 5) so that x = 4, 7.
Expanding out and cancelling the
taking the square root gives
◦
Checking both values shows that
x=7
sides gives
does not actually make the derivative
both terms in the sum are positive. So the only critical point is
◦
p
p
32+ (x − 1)2 + 36+ (x − 5)2 + 4.
2x − 2
P 0 (x) = 0 + p
+p
.
2 (x − 1)2 + 36
(x − 5)2 + 4
and squaring both sides and cross-multiplying gives
4(x − 1)2 = 36(x − 5)2 ;
P 0 (x)
equal to zero, as
x = 4.
By the properties of the target function we see that the perimeter takes its minimum at
x
•
√
◦
and this is the function
x=4
, since if
is large positive or large negative, the perimeter will be large.
Example: A at wae of area
π
square units in the shape of an isosceles triangle is to be rolled into a cone
(for ice cream). What dimensions of the triangle will t the largest volume of ice cream inside the cone?
◦
Let the base of the triangle be
b
units, the height of the triangle be
units, and the height of the cone be
1 2
πr d,
3
d
h
units, the radius of the cone be
r
units. All of these are lengths, so they are all positive.
◦
The volume of the cone is
◦
We also see that the area of the triangle is
the base of the isosceles triangle becomes
the circumference of the bottom of the
Finally, the height of the isosceles triangle
which is the function we want to maximize.
1
bh = π . Since
2
cone, we see 2πr = b.
becomes the slant height of the cone, so by the Pythagorean Theorem we see that
◦
h=
√
r 2 + d2 .
We now need to eliminate all but one variable: let's write the volume of the cone in terms of the radius
r.
r.
1
1
1
◦ We know 2πr = b so the area condition bh = π says (2πr)h = π , or πrh = π , so that h = .
2
2
r
r
√
1
1
◦ Now since h = r2 + d2 we have h2 = r2 + d2 or d2 = 2 − r2 , so d =
− r2 .
r
r2
r
√
1
1 2
◦ Plugging in shows that V (r) = πr ·
− r2 . Moving the r2 inside the square root as r4
2
3
r
1 √ 2
slightly simpler expression V (r) =
π · r − r6 .
3
To do this we need to solve for
d
in terms of
24
gives a
dV
1
2r − 6r5
= π· √
.
dr
3
2 r2 − r6
2r − 6r5 = 0, or 2r(1 − 3r4 ) = 0.
◦
Now we can dierentiate to get
◦
The derivative is zero when
r
makes this true is
◦
r=
4
1
.
3
The only value of
r
between 0 and 1 which
So there is only one critical point which could potentially be a maximum.
r ≥ 1, the function under the square root in the denominator will be zero or negative, so we ignore
r. Similarly, we can ignore r ≤ 0 since r is a length.
r
4 1
is the
the physical setup of the problem (or by the Second Derivative Test), we see that r =
3
When
these values of
◦
By
desired maximum.
r
◦
The triangle dimensions for this
r
are
b = 2πr = 2π ·
r
◦
Remark: For this optimal cone, the radius is
4
4
1
≈ 4.77 units
3
, and
√
1
4
= 3 ≈ 1.32 units
r
h=
1
≈ 0.76 units and the height is
3
r
√
.
1
3 − √ ≈ 1.07 units.
3
(Based on these dimensions, we can see that real ice cream cones are decidedly not shaped to contain
the maximal amount of ice cream!)
3.6
•
Antiderivatives, Applications to Physics
Now that we know how to dierentiate functions, we might ask whether we can reverse the process: in other
words, given a function
◦
•
For example, if
f,
is there a function
f (x) = 2x
It turns out that (provided
although
◦
g
f
g
whose derivative is
then we could take
g
Or we could take
g(x) = x2 + 2.
is continuous) the answer is yes: there is a function
might be much more complicated than
Such a function
g(x) = x2 .
f?
g
whose derivative is
f,
f.
can be found by integrating the function
f
on an appropriate interval. Since we have
not discussed integration yet, we will postpone most of the discussion on how to nd antiderivatives until
later.
•
Denition: Any function
•
We will start by analyzing the antiderivatives of the simplest possible function: the identically zero function.
•
Theorem (Zero Derivative): If
◦
F (x)
whose derivative is
f 0 (x) = 0
for every
f (x)
x
is called an antiderivative of
in an interval
I,
then
f
f.
is constant on the interval
I.
Note that any constant function has derivative zero, so in fact we see that the functions with zero
derivative are precisely the constant functions.
◦
f 0 (x) = 0 for all x
the interval [a, b].
Proof: Suppose
Theorem to
in
I.
Choose any real numbers
a<b
in
I
and apply the Mean Value
f (b) − f (a)
.
b−a
f (b) − f (a)
◦ Since f 0 (x) is identically zero on I and c is in I , we see that 0 =
.
b−a
◦ Thus, so f (b) − f (a) = 0, and so f (a) = f (b). But now since f (a) = f (b) for any
in the interval I , we conclude that f (x) must be constant on the entire interval.
◦
•
•
The theorem implies that there exists a
c
in
(a, b)
such that
f 0 (c) =
two numbers
a
and
b
Using this result we can show that the antiderivative of a function is almost unique:
F1 (x) and F2 (x) are two antiderivatives of f (x) on an interval I ,
F1 (x) = F2 (x) + C for all x in I .
Corollary: If
that
◦
This means that any two antiderivatives of
whole interval). And clearly, if
F
f
then there is a constant
dier by a constant (at least, provided
is one antiderivative of
25
f,
then
F (x) + C
f
C
such
is dened on the
is also an antiderivative of
f.
◦
◦
◦
Proof: Dene
Taking derivatives gives
Therefore,
h(x)
◦
•
h(x) = F1 (x) − F2 (x)
0
h (x) =
x
for
F10 (x)
−
in
I.
F20 (x)
= f (x) − f (x) = 0,
0
h (x) is the identically zero function on I :
h(x) = C .
x
for any
in
I.
but now by the previous theorem, we conclude that
is a constant function, say
Then
F1 (x) = F2 (x) + C
for all
x
I,
in
as desired.
One application of the derivative and antiderivative is to basic physics.
◦
Recall that if the position of an object is given by
x0 (t),
rst derivative
◦
x(t),
then the velocity of that object is given by the
and the acceleration is given by the second derivative
By Newton's second law of motion ( F
= ma), applying
x00 (t).
x00 (t).
a force to an object causes it to accelerate:
thus, a force will change the second derivative
◦
So, if we know the forces acting on an object (or equivalently, if we know the object's acceleration), along
with some starting information, then by taking antiderivatives, we can determine the object's velocity
and position.
•
Example (constant acceleration): An object moves with constant acceleration
v0
velocity
at time
Find its position
x00 (t) = v 0 (t) = a
x(t)
◦
We are given that
◦
The antiderivative of the constant function
so
◦
C = v0 ,
◦
and
and velocity
v(t)
is a constant, and also
at time
a,
starting at position
x0
with
t.
v(0) = v0 .
a is at+C , so v(t) = at+C .
Plugging in
t = 0 gives v(0) = C ,
v(t) = at + v0 .
x(t) =
Taking the antiderivative again gives
setting
•
t = 0.
t = 0: x(0) = D,
so
D = x0 .
Thus the position is given by
x(t) =
1 2
at + v0 t + D,
2
1 2
at + v0 t + x0
2
Example: A particle accelerates from rest along a line, with
and we can solve for the constant
and the velocity is given by
v(t) = at + v0
a(t) = 6t2 m/sec from t = 0 to t = 10 sec,
D
by
.
and then
travels at its constant velocity for another 10 seconds. How far has it traveled from its original position after
the total 20 seconds?
10 sec, a(t) = 6t2 .
◦
For
◦
Taking the antiderivative and noting that
◦
Taking the antiderivative again gives
◦
Therefore, after 10 seconds, the particle's
◦
Over the remaining 10 seconds, the particle moves at 2000 m/sec, so it travels
◦
The total distance traveled by the particle is therefore
t
between
0
and
v(0) = 0,
v(t) = 6 ·
we see that
t4
t4
= .
4
2
velocity is 2000 m/sec,
t3
= 2t3 .
3
x(t) = 2 ·
and its position is
25000 m
5000 m
from its start.
20000 m.
.
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2012-2016. You may not reproduce or distribute this
material without my express permission.
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