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In this section, we will learn:
How the derivative of a function gives us the direction in which the curve proceeds at each point.
1.
4
APPLICATIONS OF DIFFERENTIATION
2.
APPLICATIONS OF DIFFERENTIATION
Many applications of calculus depend
on our ability to deduce facts about
a function f from information concerning
its derivatives.
3.
APPLICATIONS OF DIFFERENTIATION
4.3
How Derivatives Affect
the Shape of a Graph
In this section, we will learn:
How the derivative of a function gives us the direction
in which the curve proceeds at each point.
4.
DERIVATIVES AND GRAPH SHAPE
As f’(x) represents the slope of the curve
y = f(x) at the point (x, f(x)), it tells us
the direction in which the curve proceeds
at each point.
Thus, it is reasonable to expect that
information about f’(x) will provide us
with information about f(x).
5.
WHAT DOES f’ SAY ABOUT f ?
To see how the derivative of f can
tell us where a function is increasing
or decreasing, look at the figure.
Increasing functions
and decreasing
functions were
defined in Section 1.1
6.
WHAT DOES f’ SAY ABOUT f ?
Between A and B and between C and D,
the tangent lines have positive slope.
So, f’(x) > 0.
7.
WHAT DOES f’ SAY ABOUT f ?
Between B and C, the tangent lines
have negative slope.
So, f’(x) < 0.
8.
WHAT DOES f’ SAY ABOUT f ?
Thus, it appears that f increases when
f’(x) is positive and decreases when f’(x)
is negative.
To prove that this
is always the case,
we use the Mean
Value Theorem.
9.
INCREASING/DECREASING TEST (I/D TEST)
a.If f’(x) > 0 on an interval, then f is
increasing on that interval.
b.If f’(x) < 0 on an interval, then f is
decreasing on that interval.
10.
I/D TEST Proof a
Let x1 and x2 be any two numbers in
the interval with x1 < x2.
According to the definition of an increasing
function, we have to show that f(x1) < f(x2).
11.
I/D TEST Proof a (Equation 1)
Since we are given that f’(x) > 0, we know that
f is differentiable on [x1, x2].
So, by the Mean Value Theorem, there is
a number c between x1 and x2 such that:
f(x2) – f(x1) = f’(c)(x2 – x1)
12.
I/D TEST Proof a and b
Now, f’(c) > 0 by assumption and x2 – x1 > 0
because x1 < x2 .
Thus, the right side of Equation 1 is positive.
So, f(x2) – f(x1) > 0 or f(x1) < f(x2)
This shows that f is increasing.
Part (b) is proved similarly.
13.
I/D TEST Example 1
Find where the function
f(x) = 3x4 – 4x3 – 12x2 + 5
is increasing and where it is decreasing.
14.
I/D TEST Example 1
f’(x) = 12x3 - 12x2 - 24x = 12x(x – 2)(x + 1)
To use the ID Test, we have to know where
f’(x) > 0 and where f’(x) < 0.
This depends on the signs of the three factors
of f’(x)—namely, 12x, x – 2, and x + 1.
15.
I/D TEST Example 1
We divide the real line into intervals
whose endpoints are the critical numbers
-1, 0, and 2 and arrange our work in
a chart.
16.
I/D TEST Example 1
A plus sign indicates the given expression
is positive.
A minus sign indicates it is negative.
The last column gives the conclusion
based on the I/D Test.
17.
I/D TEST Example 1
For instance, f’(x) < 0 for 0 < x < 2.
So, f is decreasing on (0, 2).
It would also be true to say that f is decreasing
on the closed interval.
18.
I/D TEST Example 1
The graph of f
the information
in the chart.
19.
WHAT DOES f’ SAY ABOUT f ?
Recall from Section 4.1 that, if f has a local
maximum or minimum at c, then c must be
a critical number of f (by Fermat’s Theorem).
However, not every critical number gives rise
to a maximum or a minimum.
So, we need a test that will tell us whether or not f
has a local maximum or minimum at a critical number.
20.
WHAT DOES f’ SAY ABOUT f ?
You can see from the figure that f(0) = 5 is
a local maximum value of f because f
increases on (-1, 0) and decreases on (0, 2).
In terms of derivatives,
f’(x) > 0 for -1 < x < 0
and f’(x) < 0 for 0 < x < 2.
21.
WHAT DOES f’ SAY ABOUT f ?
In other words, the sign of f’(x)
changes from positive to negative at 0.
This observation is the basis of the following test.
22.
FIRST DERIVATIVE TEST
Suppose that c is a critical number of
a continuous function f.
a. If f’ changes from
positive to negative
at c, then f has
a local maximum at c.
23.
FIRST DERIVATIVE TEST
b. If f’ changes from negative to
positive at c, then f has a local minimum
at c.
24.
FIRST DERIVATIVE TEST
c. If f’ does not change sign at c—for example,
if f’ is positive on both sides of c or negative
on both sides—then f has no local maximum
or minimum at c.
25.
FIRST DERIVATIVE TEST
The First Derivative Test is a consequence
of the I/D Test.
For instance, in (a), since the sign of f’(x) changes
from positive to negative at c, f is increasing to the left
of c and decreasing to the right of c.
It follows that f has
a local maximum at c.
26.
FIRST DERIVATIVE TEST
It is easy to remember the test by
visualizing diagrams.
27.
WHAT DOES f’ SAY ABOUT f ? Example 2
Find the local minimum and
maximum values of the function f
in Example 1.
28.
WHAT DOES f’ SAY ABOUT f ? Example 2
From the chart in the solution to Example 1,
we see that f’(x) changes from negative to
positive at -1.
So, f(-1) = 0 is a local minimum value by
the First Derivative Test.
29.
WHAT DOES f’ SAY ABOUT f ? Example 2
Similarly, f’ changes from negative to
positive at 2.
So, f(2) = -27 is also a local minimum value.
30.
WHAT DOES f’ SAY ABOUT f ? Example 2
As previously noted, f(0) = 5 is
a local maximum value because f’(x)
changes from positive to negative at 0.
31.
WHAT DOES f’ SAY ABOUT f ? Example 3
Find the local maximum and minimum
values of the function
g(x) = x + 2 sin x 0 ≤ x ≤ 2π
32.
WHAT DOES f’ SAY ABOUT f ? Example 3
To find the critical numbers of g,
we differentiate:
g’(x) = 1 + 2 cos x
So, g’(x) = 0 when cos x = - ½.
The solutions of this equation are 2π/3 and 4π/3.
33.
WHAT DOES f’ SAY ABOUT f ? Example 3
As g is differentiable everywhere,
the only critical numbers are 2π/3 and 4π/3.
So, we analyze g in the following table.
34.
WHAT DOES f’ SAY ABOUT f ? Example 3
As g’(x) changes from positive to negative
at 2π/3, the First Derivative Test tells us
that there is a local maximum at 2π/3.
The local maximum value is:
2π 2π 2π ⎛ 3 ⎞ 2π
g (2π / 3) = + 2sin = + 2 ⎜⎜ ⎟⎟ = + 3
3 3 3 ⎝ 2 ⎠ 3
≈3.83
35.
WHAT DOES f’ SAY ABOUT f ? Example 3
Likewise, g’(x) changes from negative to
positive at 4π/3.
So, a local minimum value is:
4π 4π 4π ⎛ 3 ⎞ 4π
g (4π / 3) = + 2sin = + 2⎜
⎜ − ⎟⎟ = − 3
3 3 3 ⎝ 2 ⎠ 3
≈2.46
36.
WHAT DOES f’ SAY ABOUT f ? Example 3
The graph of g supports our
37.
WHAT DOES f’’ SAY ABOUT f ?
The figure shows the graphs of
two increasing functions on (a, b).
38.
WHAT DOES f’’ SAY ABOUT f ?
Both graphs join point A to point B, but
they look different because they bend in
different directions.
How can we distinguish between these two types
of behavior?
39.
WHAT DOES f’’ SAY ABOUT f ?
Here, tangents to these curves have
been drawn at several points.
40.
CONCAVE UPWARD
In the first figure, the curve lies
above the tangents and f is called
concave upward on (a, b).
41.
CONCAVE DOWNWARD
In the second figure, the curve lies
below the tangents and g is called
concave downward on (a, b).
42.
If the graph of f lies above all of
its tangents on an interval I, it is called
concave upward on I.
If the graph of f lies below all of its tangents
on I, it is called concave downward on I.
43.
The figure shows the graph of a function that
is concave upward (CU) on the intervals (b, c),
(d, e), and (e, p) and concave downward (CD)
on the intervals (a, b), (c, d), and (p, q).
44.
Let’s see how the second derivative
helps determine the intervals of
45.
From this figure, you can see that, going from
left to right, the slope of the tangent increases.
This means that the derivative f’ is an increasing
function and therefore its derivative f” is positive.
46.
Likewise, in this figure, the slope of
the tangent decreases from left to right.
So, f’ decreases and therefore f’’ is negative.
This reasoning can be
reversed and suggests
that the following
theorem is true.
47.
CONCAVITY TEST
a.If f’’(x) > 0 for all x in I, then the graph of f
is concave upward on I.
b.If f’’(x) < 0 for all x in I, then the graph of f
is concave downward on I.
48.
CONCAVITY Example 4
The figure shows a population graph for
Cyprian honeybees raised in an apiary.
How does the rate of population increase change
over time?
When is this rate highest?
Over what
intervals is P
concave upward or
concave downward?
49.
CONCAVITY Example 4
By looking at the slope of the curve as t
increases, we see that the rate of increase
of the population is initially very small.
Then, it gets larger
until it reaches
a maximum at
about t = 12 weeks,
and decreases as
the population
begins to level off.
50.
CONCAVITY Example 4
As the population approaches its
maximum value of about 75,000 (called
the carrying capacity), the rate of increase,
P’(t), approaches 0.
The curve
appears to be
concave upward
on (0, 12) and
concave downward
on (12, 18).
51.
INFLECTION POINT
In the example, the curve changed from
concave upward to concave downward at
approximately the point (12, 38,000).
This point is called
an inflection point
of the curve.
52.
INFLECTION POINT
The significance of this point is that
the rate of population increase has its
maximum value there.
In general, an inflection point is a point where
a curve changes its direction of concavity.
53.
INFLECTION POINT—DEFINITION
A point P on a curve y = f(x) is called
an inflection point if f is continuous there
and the curve changes from concave upward
to concave downward or from concave
downward to concave upward at P.
54.
INFLECTION POINT
For instance, here, B, C, D, and P are
the points of inflection.
Notice that, if a curve has a tangent at a point of
inflection, then the curve crosses its tangent there.
55.
INFLECTION POINT
In view of the Concavity Test, there is
a point of inflection at any point where
the second derivative changes sign.
56.
WHAT DOES f’’ SAY ABOUT f ? Example 5
Sketch a possible graph of a function f
that satisfies the following conditions:
(i) f’(x) > 0 on (-∞ , 1), f’(x) < 0 on (1, ∞)
(ii) f’’(x) > 0 on (-∞, -2) and (2, ∞), f’’(x) < 0 on (-2, 2)
(iii) ,lim f ( x) = −2 lim f ( x) = 0
x → −∞ x→ ∞
57.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition i
The first condition tells us that f
is increasing on (-∞ , 1) and decreasing
on (1, ∞).
58.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition ii
The second condition says that f
is concave upward on (-∞, -2) and (2, ∞),
and concave downward on (-2, 2).
59.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition iii
From the third condition, we know
that the graph of f has two horizontal
y = -2
y=0
60.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition iii
We first draw the horizontal asymptote
y = -2 as a dashed line.
We then draw the graph of f approaching this asymptote
at the far left—increasing to its maximum point at x = 1
and decreasing toward the x-axis at the far right.
61.
WHAT DOES f’’ SAY ABOUT f ? E. g. 5—Condition iii
We also make sure that the graph has
inflection points when x = -2 and 2.
Notice that we made the curve bend upward for x < -2
and x > 2, and bend downward when x is between -2
and 2.
62.
WHAT DOES f’’ SAY ABOUT f ?
Another application of the second
derivative is the following test
for maximum and minimum values.
It is a consequence of the Concavity Test.
63.
SECOND DERIVATIVE TEST
Suppose f’’ is continuous near c.
a.If f’(c) = 0 and f’’(c) > 0, then f has
a local minimum at c.
b.If f’(c) = 0 and f’’(c) < 0, then f has
a local maximum at c.
64.
SECOND DERIVATIVE TEST
For instance, (a) is true because f’’(x) > 0
near c, and so f is concave upward near c.
This means that
the graph of f lies
above its horizontal
tangent at c, and so
f has a local minimum
at c.
65.
WHAT DOES f’’ SAY ABOUT f ? Example 6
Discuss the curve
y = x4 – 4x3
with respect to concavity, points of inflection,
and local maxima and minima.
Use this information to sketch the curve.
66.
WHAT DOES f’’ SAY ABOUT f ? Example 6
If f(x) = x4 – 4x3, then:
f’(x) = 4x3 – 12x2 = 4x2(x – 3)
f’’(x) = 12x2 – 24x = 12x(x – 2)
67.
WHAT DOES f’’ SAY ABOUT f ? Example 6
To find the critical numbers, we set f’(x) = 0
and obtain x = 0 and x = 3.
To use the Second Derivative Test,
we evaluate f’’ at these critical numbers:
f’’(0) = 0 f’’(3) = 36 > 0
68.
WHAT DOES f’’ SAY ABOUT f ? Example 6
As f’(3) = 0 and f’’(3) > 0, f(3) = -27 is
a local minimum.
As f’’(0) = 0, the Second Derivative Test gives
no information about the critical number 0.
69.
WHAT DOES f’’ SAY ABOUT f ? Example 6
However, since f’(x) < 0 for x < 0 and also
for 0 < x < 3, the First Derivative Test tells
us that f does not have a local maximum or
minimum at 0.
In fact, the expression for f’(x) shows that
f decreases to the left of 3 and increases to
the right of 3.
70.
WHAT DOES f’’ SAY ABOUT f ? Example 6
As f’’(x) = 0 when x = 0 or 2, we divide
the real line into intervals with those numbers
as endpoints and complete the following chart.
71.
WHAT DOES f’’ SAY ABOUT f ? Example 6
The point (0, 0) is an inflection point—since
the curve changes from concave upward to
concave downward there.
72.
WHAT DOES f’’ SAY ABOUT f ? Example 6
Also, (2, -16) is an inflection point—since
the curve changes from concave downward
to concave upward there.
73.
WHAT DOES f’’ SAY ABOUT f ? Example 6
Using the local minimum, the intervals
of concavity, and the inflection points,
we sketch the curve.
74.
The Second Derivative Test is
inconclusive when f’’(c) = 0.
In other words, at such a point, there might be
a maximum, a minimum, or neither (as in the example).
75.
The test also fails when f’’(c) does not exist.
In such cases, the First Derivative Test
must be used.
In fact, even when both tests apply, the First Derivative
Test is often the easier one to use.
76.
WHAT DOES f’’ SAY ABOUT f ? Example 7
Sketch the graph of the function
f(x) = x2/3(6 – x)1/3
You can use the differentiation rules to check that
the first two derivatives are:
4 −x −8
f '( x) = 1/ 3 f ''( x) = 4 / 3
x (6 −x) 2 / 3 x (6 −x)5 / 3
As f’(x) = 0 when x = 4 and f’(x) does not exist when
x = 0 or x = 6, the critical numbers are 0, 4, and 6.
77.
WHAT DOES f’’ SAY ABOUT f ? Example 7
To find the local extreme values, we
use the First Derivative Test.
As f’ changes from negative to positive at 0,
f(0) = 0 is a local minimum.
78.
WHAT DOES f’’ SAY ABOUT f ? Example 7
Since f’ changes from positive to negative at 4,
f(4) = 25/3 is a local maximum.
The sign of f’ does not change at 6, so there is
no minimum or maximum there.
79.
WHAT DOES f’’ SAY ABOUT f ? Example 7
The Second Derivative Test could be
used at 4, but not at 0 or 6—since f’’ does
not exist at either of these numbers.
80.
WHAT DOES f’’ SAY ABOUT f ? Example 7
Looking at the expression for f’’(x)
and noting that x4/3 ≥ 0 for all x,
we have:
f’’(x) < 0 for x < 0 and for 0 < x < 6
f’’(x) > 0 for x > 6
81.
WHAT DOES f’’ SAY ABOUT f ? Example 7
So, f is concave downward on (-∞, 0) and
(0, 6) and concave upward on (6, ∞), and
the only inflection point is (6, 0).
Note that the curve
has vertical tangents
at (0, 0) and (6, 0)
because |f’(x)| → ∞
as x → 0 and as x → 6.
82.
WHAT DOES f’’ SAY ABOUT f ? Example 8
Use the first and second derivatives
of f(x) = e1/x, together with asymptotes,
to sketch its graph.
Notice that the domain of f is {x | x ≠ 0}.
So, we check for vertical asymptotes by
computing the left and right limits as x → 0.
83.
WHAT DOES f’’ SAY ABOUT f ?
As x → 0+, we know that t = 1/x → ∞.
1/ x t
So, lim+ e = lim e = ∞
x→ 0 t→ ∞
This shows that x = 0 is a vertical asymptote.
84.
WHAT DOES f’’ SAY ABOUT f ?
As x → 0-, we know that t = 1/x → -∞.
1/ x t
So, lim− e = lim e = 0
t → −∞
x→ 0
85.
WHAT DOES f’’ SAY ABOUT f ?
As x → ±∞, we have 1/x → 0.
1/ x 0
So, lim e = e =1
x → ±∞
This shows that y = 1 is a horizontal asymptote.
86.
WHAT DOES f’’ SAY ABOUT f ?
Now, let’s compute the derivative.
1/ x
e
The Chain Rule gives: f '( x ) = − 2
x
Since e1/x > 0 and x2 > 0 for all x ≠ 0, we have
f’(x) < 0 for all x ≠ 0.
Thus, f is decreasing on (-∞, 0) and on (0, ∞) .
87.
WHAT DOES f’’ SAY ABOUT f ?
There is no critical number.
So, the function has no maximum
or minimum.
88.
WHAT DOES f’’ SAY ABOUT f ?
The second derivative is:
2 1/ x 2 1/ x
x e (−1/ x ) −e (2 x)
f ''( x) = − 4
x
1/ x
e (2 x + 1)
= 4
x
89.
WHAT DOES f’’ SAY ABOUT f ?
As e1/x > 0 and x4 > 0, we have:
f’’(x) > 0 when x > -½ (x ≠ 0)
f’’(x) < 0 when x < -½
So, the curve is concave downward on (-∞, -½)
and concave upward on (-½, 0) and on (0, ∞).
The inflection point is (-½, e-2).
90.
WHAT DOES f’’ SAY ABOUT f ?
To sketch the graph of f, we first draw
the horizontal asymptote y = 1 (as a dashed
line), together with the parts of the curve
near the asymptotes
in a preliminary sketch.
91.
WHAT DOES f’’ SAY ABOUT f ?
These parts reflect the information concerning
limits and the fact that f is decreasing on both
(-∞, 0) and (0, ∞).
Notice that we have
indicated that f(x) → 0
as x → 0- even though
f(0) does not exist.
92.
WHAT DOES f’’ SAY ABOUT f ?
Here, we finish the sketch by
incorporating the information concerning
concavity and the inflection point.
93.
WHAT DOES f’’ SAY ABOUT f ?
Finally, we check our work with
a graphing device.
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