Transcript f ``(x)

Graphs and the Derivative
Chapter 13
Ch. 13 Graphs and the Derivative



13.1 Increasing and Decreasing Functions
13.2 Relative Extrema
13.3 Higher Derivatives, Concavity, and the
Second Derivative Test
13.1 Increasing and Decreasing Functions


The graph of a typical function may increase on some
intervals and decrease on others.
The derivative of a function f ’(x) describes whether the
graph is increasing (rising from left to right) or
decreasing (falling from left to right).
 Where f ’(x) > 0, f(x) will increase.
 Where f ’(x) < 0, f(x) will decrease.
 Where f ’(x) = 0, f(x) is constant.
Increasing and Decreasing Functions

f(x)
f ’(x) > 0, f(x) Increasing
f ’(x) = 0,
f(x) is constant
f ’(x) < 0, f(x) Decreasing
Increasing and Decreasing Functions
CRITICAL NUMBERS
 The critical numbers for a function f are those numbers c
in the domain of f for which f ’(c) = 0 or f ’(c) does not
exist.
CRITICAL POINT
 A critical point is a point whose x-coordinate is the
critical number c, and whose y-coordinate is f(c).
Increasing and Decreasing Functions
f(c)
critical numbers c
c
f ’(c) = 0
Increasing and Decreasing Functions
f(c)
c
critical points [c, f(c)]
Increasing and Decreasing Functions

Test for increasing or decreasing functions
1. Determine the critical numbers
 Calculate the derivative of f(x)
 Set f ’(x) equal to 0
2. Locate the critical numbers on a number line to
identify “intervals”.
3. Choose a value of x in each interval to determine if
f ’(x) > 0 or < 0 in that interval.
4. Use the test above to determine if f is increasing or
decreasing in that interval.
Increasing and Decreasing Functions

Example: f(x) = x3 + 3x2 – 9x + 4
1. Determine the critical numbers
 f ’(x) = 3x2 + 6x – 9
 3x2 + 6x – 9 = 0
 3(x + 3)(x – 1) = 0
 x = -3
or x = 1 (critical numbers)
2. Locate the critical numbers on a number line to
identify “intervals”.
f(x) = x3 + 3x2 – 9x + 4
f ’(x) = 3x2 + 6x – 9
35
30
25
f(x)
20
15
3 intervals, (- , -3),10(-3, 1), (1, )
5
0
-6
-5
-4
-3
-2
-1 -5 0
x
1
2
3
4
Increasing and Decreasing Functions

Example: f(x) = x3 + 3x2 – 9x + 4
1. Determine the critical numbers
 f ’(x) = 3x2 + 6x – 9
 3x2 + 6x – 9 = 0
x = -3 or x = 1
2. Locate the critical numbers on a number line to
identify “intervals”.
3. Choose a value of x in each interval to determine if
f ’(x) > 0 or < 0 in that interval.
4. Use the test above to determine if f is increasing or
decreasing in that interval.
f(x) = x3 + 3x2 – 9x + 4
f ’(x) = 3x2 + 6x – 9
f(x)
35
f ’(-4) = 3(-4)2 + 6(-4) – 9 = 15
30
f ’(x) > 0. f(x) is increasing
25
f ’(0) = -9
20
f ’ (x) < 0. f(x) is decreasing
15 f ’(2) = 15
10 f ’ (x) > 0. f(x) is increasing
5
0
-6
-5
-4
-3
-2
-1 -5 0
x
1
2
3
4
f(x) = x3 + 3x2 – 9x + 4
f ’(x) = 3x2 + 6x – 9
35
30
25
f(x)
20
15
10
5
0
-6
-5
-4
-3
-2
-1 -5 0
x
1
2
3
4
Now You Try

A manufacturer sells video games with the following
cost and revenue functions, where x is the number of
games sold.
2
C  x   4.8 x  .0004 x , 0  x  2250
R  x   8.4 x  .002 x 2 ,

0  x  2250
Determine the interval(s) on which the profit function is
increasing.
Remember: P(x) = R(x) – C(x)
Relative Extrema
RELATIVE MAXIMUM OR MINIMUM
Let c be a number in the domain of a function f. Then
 f(c) is a relative maximum for f if there exists an open
interval (a, b) containing c such that
f(x)  f(c)
for all x in (a, b)
 f(c) is a relative minimum for f if there exists an open
interval (a, b) containing c such that
f(x)  f(c)
for all x in (a, b)
 A function has a relative extremum at c if it has either a
relative maximum or relative minimum there.
Relative Extrema
If a function f has a relative extremum at c, then c is a
critical number or c is an endpoint of the domain.
relative
maximum
35
•
30
25
Increasing
20
f(x)

Increasing
15
Decreasing 10
5
0
-6
-5
-4
-3
-2
-1 -5 0
x
•
1
2
relative
minimum
3
4
Relative Extrema
30
Critical point (f ’(x) = 0), but
NOT a relative extremum
20
Increasing
f(x)
10
0
-4
-2
0
2
4
f  x   x3
-10
Increasing
-20
-30
x
Relative Extrema - FIRST DERIVATIVE TEST

Let c be a critical number for a function f. Then, the
critical point (c, f(c)) is
 A relative maximum if f ’(x) > 0 to the left of c and
f ’(x) < 0 to the right of c
 A relative minimum if f ’(x) < 0 to the left of c and
f ’(x) > 0 to the right of c
 Not a relative extrema if f ’(x) has the same sign on
both sides of c.
Relative Extrema
relative
maximum
35
•
30
25
f ’(x) > 0
f(x)
20
f ’(x) > 0
15
f ’(x) < 0
10
5
0
-6
-5
-4
-3
-2
-1 -5 0
x
•
1
2
relative
minimum
3
4
Relative Extrema
30
Critical point (f ’(x) = 0), but
NOT a relative extremum
20
f ’(x) > 0
f(x)
10
0
-4
-2
0
2
4
f  x   x3
-10
f ’(x) > 0
-20
-30
x
Now You Try


a.
b.
The total profit P(x) (in thousands of dollars) from the
sale of x units of a certain prescription drug is given by
P(x) = ln(-x3 + 3x2 + 72x + 1)
for x in [0, 10].
Find the number of units that should be sold in order to
maximize the total profit.
What is the maximum profit?
Try Another

The number of passenger cars imported into the United
States (in millions) in year x can be approximated by
f(x) = –.051x3 + 1.43x2 – 12.55x + 39.66

(0  x  13),
where x = 5 corresponds to 1995. Find all relative
extrema of this function and interpret your answers.
13.3 Higher Derivatives


If a function f has a derivative f ’, then the derivative of f ’
is the second derivative of f, written f ’’.
f ’’’ is the third derivative of f, and so on.
f  x   x4  2x3  3x2  5x  7
f '  x   4 x3  6 x 2  6 x  5
f ''  x   12x2  12x  6
f '''  x   24 x  12
f  4  x   24
Higher Derivatives
NOTATION FOR HIGHER DERIVATIVES
 The second derivative of y = f(x) can be written using
any of the following notations:
2
f ''( x),

d y
,
2
dx
Dx2  f  x  
For n  4, the nth derivative is written f(n)(x).
The second derivative f ’’(x) is the rate of change of the
first derivative f ’(x).
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Total Product
P(L)
Quantity of Labor
Total product is
increasing,
marginal
product > 0
P’(L)
Quantity of Labor
Marginal
Product
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Total Product
P(L)
Quantity of Labor
P’(L)
Quantity of Labor
Marginal
Product
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Total Product
P(L)
Quantity of Labor
Increasing
Marginal
Returns
P’(L)
Quantity of Labor
Marginal
Product
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Total Product
P(L)
Quantity of Labor
Diminishing
Marginal
Returns
P’(L)
Quantity of Labor
Marginal
Product
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Total Product
P(L)
Quantity of Labor
Negative
Marginal
Returns
P’(L)
Quantity of Labor
Marginal
Product
Second Derivative
P ' L  0
P ''( L)  0
P’(L)
L
P ' L  0
P ' L  0
P ''( L)  0
P ''( L)  0
Marginal
Product
Now You Try

Find f ’’(x). Then find f ’’(0) and f ’’(2).
f  x    x 4  2 x3  x 2
Concavity of a Graph



First derivative shows...
a. where a function f is increasing or decreasing and,
b. where the extrema occur.
Second derivative gives the rate of change of the first
derivative, or
 how fast the function is increasing or decreasing
Rate of change of the derivative (f ’’(x)) affects the shape
of the graph.
 Concave upward
 Concave downward
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Concave
upward
Total Product
P(L)
Concave
downward
Quantity of Labor
P’(L)
Quantity of Labor
Marginal
Product
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Concave
upward
Total Product
P(L)
Concave
downward
Quantity of Labor
P’(L)
P ''( L)  0
P ''( L)  0
Quantity of Labor
Marginal
Product
Concavity of a Graph
TEST FOR CONCAVITY
 Let f be a function with derivatives f ’ and f ’’ existing
for all points in an interval (a, b).
1. If f ’’(x) > 0 for all x in (a, b), then f is concave upward on
(a, b).
2. If f ’’(x) < 0 for all x in (a, b), then f is concave downward
on (a, b).
Point of Inflection: Point where a graph changes
concavity (f ’’(x) = 0 ).
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Concave
upward
Total Product
P(L)
Concave
downward
Point of Inflection
P ''( L)  0
Quantity of Labor
m=0
P’(L)
Quantity of Labor
Marginal
Product
Concavity of a Graph
TEST FOR CONCAVITY
 Let f be a function with derivatives f ’ and f ’’ existing
for all points in an interval (a, b).
1. If f ’’(x) > 0 for all x in (a, b), then f is concave upward on
(a, b).
2. If f ’’(x) < 0 for all x in (a, b), then f is concave downward
on (a, b).
Point of Inflection: Point where a graph changes
concavity (f ’’(x) = 0 ).
Find all intervals where f(x) = x4 – 8x3 + 18x2 is concave
upward or downward, and find all inflection points.
Concavity of a Graph
f  x   x4  8x3  18x2
f '  x   4 x3  24x2  36x
f ''  x   12x2  48x  36
f ''  x   12  x 1 x  3
f '' 1  0, f '' 3  0
3 intervals: (-, 1), (1, 3), (3, )
Test each interval for concavity:
f ''  0   12  0   48  0   36  36 f concave upward on (-, 1)
2
f ''  2  12 f concave downward on (1, 3)
f ''  4  36 f concave upward on (3, )
Concavity of a Graph
Inflection points at x = 1 and x = 3
f 1  1  8 1  18 1  11
4
3
2
f  3   3  8  3  18  3  27
4
3
2
Points of inflection: (1, 11) and (3, 27)
f  x   x4  8x3  18x2
Concavity of a Graph
80
70
60
50
f(x)
40
30
20
(1, 11)
10
0
-2
-1
-10 0
1
2
x
Concave upward
f ’’(x) > 0
3
4
5
6
f  x   x4  8x3  18x2
Concavity of a Graph
80
70
60
50
f(x)
40
(3, 27)
30
(1, 11)
20
10
0
-2
-1
-10 0
1
2
x
3
4
Concave downward
f ’’(x) < 0
5
6
f  x   x4  8x3  18x2
Concavity of a Graph
80
70
60
50
f(x)
40
(3, 27)
30
20
(1, 11)
10
0
-2
-1
-10 0
1
2
3
4
5
6
x
Concave upward
f ’’(x) > 0
Point of Diminishing Returns


Law of Diminishing Returns – The principle that as
additional units of a variable resource (labor) are added
to a fixed resource (capital), the marginal product will
eventually decrease.
Point of Diminishing Returns – The point on the
production function at which the marginal product
begins to decrease as additional units of a variable
resource are added to a fixed resource.
 Occurs at the point of inflection where concavity
changes from upward to downward.
Short-Run Production Relationships
Law of Diminishing Marginal Returns
Total Product
P(L)
Point of Diminishing
Marginal Returns
Quantity of Labor
P’(L)
Quantity of Labor
Marginal
Product
Point of Diminishing Returns - example
An efficiency study of the morning shift (8:00 – noon) at
a factory indicates that an average worker who starts at
8:00 A.M. will have produced Q(t) = -t3 + 9t2 + 12t units
t hours later. At what time during the shift is the worker
performing most efficiently (point of diminishing
returns)?
 Solution:
The worker’s rate of production is the derivative of Q(t)

R(t) = Q’(t) = -3t2 + 18t + 12

The point of diminishing returns occurs at a point of
inflection where Q’’(t) = 0 for 0  t  4
Q’’(t) = -6t + 18
Point of Diminishing Returns - example

Q’’(t) = -6t + 18 = 0 when t = 3,
 is

positive for 0 < t < 3 (concave upward),
 and negative for 3 < t < 4 (concave downward).
Therefore, maximum efficiency occurs at t = 3, or at
11:00.
Point of Diminishing Returns - example
140.00
Q(t) = -t3 + 9t2 + 12t
120.00
Q(t)
100.00
80.00
60.00
40.00
20.00
0.00
0.00
1.00
2.00
3.00
t
4.00
5.00
Point of Diminishing Returns - example
140.00
Q(t) = -t3 + 9t2 + 12t
120.00
Q(t)
100.00
Point of inflection
(3.00, 90)
80.00
60.00
40.00
20.00
0.00
0.00
1.00
2.00
3.00
t
4.00
5.00
Point of Diminishing Returns - example
Q’(t) = -3t2 + 18t + 12
45.00
Q’’(t) = 0
40.00
35.00
Q'(t )
30.00
25.00
20.00
15.00
10.00
5.00
0.00
0.00
1.00
2.00
3.00
t
4.00
5.00
45.00
40.00
35.00
Q'(t )
30.00
25.00
20.00
15.00
10.00
5.00
0.00
0.00
1.00
2.00
3.00
4.00
5.00
3.00
4.00
5.00
t
140.00
120.00
Q(t)
100.00
80.00
60.00
40.00
20.00
0.00
0.00
1.00
2.00
t
Concavity of a Graph
SECOND DERIVATIVE TEST FOR RELATIVE EXTREMA
Let f ’’ exist on some open interval containing c, and
let f ‘(c) = 0
1. If f ‘’(c) > 0, then f has a relative minimum at c.
f(c) is concave upward on a < c < b.
c, f(c)
Concavity of a Graph
SECOND DERIVATIVE TEST FOR RELATIVE EXTREMA
Let f ’’ exist on some open interval containing c, and
let f ‘(c) = 0
1. If f ‘’(c) > 0, then f has a relative minimum at c.
2. If f ‘’(c) < 0, then f has a relative maximum at c.
f(c) is concave downward on a < c < b.
c, f(c)
Concavity of a Graph
SECOND DERIVATIVE TEST FOR RELATIVE EXTREMA
Let f ’’ exist on some open interval containing c, and
let f ‘(c) = 0
1. If f ‘’(c) > 0, then f has a relative minimum at c.
2. If f ‘’(c) < 0, then f has a relative maximum at c.
3. If f ‘’(c) = 0, then the test gives no information about
extrema; use the first derivative test.
A Previous Exercise

The number of passenger cars imported into the United
States (in millions) in year x can be approximated by
f(x) = –.051x3 + 1.43x2 – 12.55x + 39.66

(0  x  13),
where x = 5 corresponds to 1995. Find all relative
extrema of this function and interpret your answers.
f '  x   .153x2  2.86x 12.55
f '  0  7.0 and 11.7
A Previous Exercise
f '  x   .153x  2.86x 12.55
2
f '  0  7.0 and 11.7
f " x   .306x  2.86
f " 7.0  .306  7.0  2.86  .718  0
There is a relative minimum at x = 7.0
f "11.7   .306 11.7   2.86  .7202  0
There is a relative maximum at x = 11.7
A Previous Exercise
f(x) = –.051x3 + 1.43x2 – 12.55x + 39.66
8
7
imports
6
(11.7, 6.90)
5
4
(7, 4.387)
3
2
1
0
5
6
7
8
9
10
year
11
12
13
14
A Previous Exercise
To determine points of inflection:
 Set f “(x) equal to zero and solve for x
f " x   .306x  2.86
f " 0  9.3

Calculate f (9.3)
f ( x)  .051x3  1.43x 2  12.55x  39.66
f (9.3)  .051 9.3  1.43  9.3  12.55  9.3  39.66
3
 5.6

Point of inflection : (9.3, 5.6)
2
A Previous Exercise
f(x) = –.051x3 + 1.43x2 – 12.55x + 39.66
8
7
(9.3, 5.6)
imports
6
(11.7, 6.90)
5
4
(7, 4.387)
3
2
1
0
5
6
7
8
9
10
year
11
12
13
14
Now You Try

Find any critical numbers for f and then use the second
derivative test to decide whether the critical numbers are
relative maxima or relative minima.
f  x   3x 3  3x 2  1
Chapter 13
End