Transcript The x

4
Applications of the Derivative
 Applications of the First Derivative
 Applications of the Second Derivative
 Curve Sketching
 Optimization I
 Optimization II
4.1
Applications of the First Derivative
Relative
Maximum
60
y
f ( x)
40
20
–7 –5
–3
–1
1
3
5
–20
–40
–60
Relative
Minimum
7
x
Increasing and Decreasing Functions
 A function f is increasing on an interval (a, b)
if for any two numbers x1 and x2 in (a, b),
f(x1) < f(x2) wherever x1 < x2.
y
f(x2)
f(x1)
x
a
x1
x2
b
Increasing and Decreasing Functions
 A function f is decreasing on an interval (a, b)
if for any two numbers x1 and x2 in (a, b),
f(x1) > f(x2) wherever x1 < x2.
y
f(x1)
f(x2)
x
a
x1
x2
b
Theorem 1
 If f ′(x) > 0 for each value of x in an interval (a, b),
then f is increasing on (a, b).
 If f ′(x) < 0 for each value of x in an interval (a, b),
then f is decreasing on (a, b).
 If f ′(x) = 0 for each value of x in an interval (a, b),
then f is constant on (a, b).
Example
 Find the interval where the function f(x) = x2 is increasing
and the interval where it is decreasing.
Solution
 The derivative of f(x) = x2
is f ′(x) = 2x.
 f ′(x) = 2x > 0 if x > 0
and f ′(x) = 2x < 0 if x < 0.
 Thus, f is increasing on the
interval (0, ) and decreasing
on the interval (– , 0).
y
f(x) = x2
5
4
3
2
1
–2
Example 1, page 245
–1
0
1
2
x
Determining the Intervals Where a Function
is Increasing or Decreasing
1. Find all the values of x for which f ′(x) = 0 or f ′ is
discontinuous and identify the open intervals
determined by these numbers.
2. Select a test number c in each interval found in step 1
and determine the sign of f ′(c) in that interval.
a. If f ′(c) > 0, f is increasing on that interval.
b. If f ′(c) < 0, f is decreasing on that interval.
Examples
 Determine the intervals where the function
f(x) = x3 – 3x2 – 24x + 32
is increasing and where it is decreasing.
Solution
1. Find f ′ and solve for f ′(x) = 0:
f ′(x) = 3x2 – 6x – 24 = 3(x + 2)(x – 4) = 0
✦ Thus, the zeros of f ′ are x = –2 and x = 4.
✦ These numbers divide the real line into the intervals
(– , –2), (– 2, 4), and (4, ).
Example 2, page 246
Examples
 Determine the intervals where the function
f(x) = x3 – 3x2 – 24x + 32
is increasing and where it is decreasing.
Solution
2. To determine the sign of f ′(x) in the intervals we found
(– , –2), (– 2, 4), and (4, ), we compute f ′(c) at a
convenient test point in each interval.
✦ Lets consider the values –3, 0, and 5:
f ′(–3) = 3(–3)2 – 6(–3) – 24 = 27 +18 – 24 = 21 > 0
f ′(0) = 3(0)2 – 6(0) – 24 = 0 +0 – 24 = –24 < 0
f ′(5) = 3(5)2 – 6(5) – 24 = 75 – 30 – 24 = 21 > 0
✦ Thus, we conclude that f is increasing on the intervals
(– , –2), (4, ), and is decreasing on the interval (– 2, 4).
Example 2, page 246
Examples
 Determine the intervals where the function
f(x) = x3 – 3x2 – 24x + 32
is increasing and where it is decreasing.
Solution
 So, f increases on (– , –2), (4, ), and decreases on (– 2, 4):
60
y
y = x3 – 3x2 – 24x + 32
40
20
–7 –5
–3
–1
–20
–40
–60
Example 2, page 246
1
3
5
7
x
Examples
 Determine the intervals where f ( x )  x 
and where it is decreasing.
Solution
1. Find f ′ and solve for f ′(x) = 0:
1
is increasing
x
1 x2  1
f ( x )  1  2 
0
2
x
x
✦ f ′(x) = 0 when the numerator is equal to zero, so:
x2  1  0
x2  1
x  1
✦ Thus, the zeros of f ′ are x = –1 and x = 1.
✦ Also note that f ′ is not defined at x = 0, so we have four
intervals to consider: (– , –1), (– 1, 0), (0, 1), and (1, ).
Example 4, page 247
Examples
1
 Determine the intervals where f ( x )  x 
is increasing
x
and where it is decreasing.
Solution
2. To determine the sign of f ′(x) in the intervals we found
(– , –1), (– 1, 0), (0, 1), and (1, ), we compute f ′(c) at a
convenient test point in each interval.
✦ Lets consider the values –2, –1/2, 1/2, and 2:
f ( 2)  1 
1
 2 
2
 1
1 3
 0
4 4
✦ So f is increasing in the interval (– , –1).
Example 4, page 247
Examples
1
 Determine the intervals where f ( x )  x 
is increasing
x
and where it is decreasing.
Solution
2. To determine the sign of f ′(x) in the intervals we found
(– , –1), (– 1, 0), (0, 1), and (1, ), we compute f ′(c) at a
convenient test point in each interval.
✦ Lets consider the values –2, –1/2, 1/2, and 2:
f (  12 )  1 
1
 
1 2
2
 1
1
1
4
 1  4  3  0
✦ So f is decreasing in the interval (– 1, 0).
Example 4, page 247
Examples
1
 Determine the intervals where f ( x )  x 
is increasing
x
and where it is decreasing.
Solution
2. To determine the sign of f ′(x) in the intervals we found
(– , –1), (– 1, 0), (0, 1), and (1, ), we compute f ′(c) at a
convenient test point in each interval.
✦ Lets consider the values –2, –1/2, 1/2, and 2:
f ( 12 )  1 
1
 
1 2
2
 1
1
1
4
 1  4  3  0
✦ So f is decreasing in the interval (0, 1).
Example 4, page 247
Examples
1
 Determine the intervals where f ( x )  x 
is increasing
x
and where it is decreasing.
Solution
2. To determine the sign of f ′(x) in the intervals we found
(– , –1), (– 1, 0), (0, 1), and (1, ), we compute f ′(c) at a
convenient test point in each interval.
✦ Lets consider the values –2, –1/2, 1/2, and 2:
f (2)  1 
1
 2
2
1 3
 1   0
4 4
✦ So f is increasing in the interval (1, ).
Example 4, page 247
Examples
1
 Determine the intervals where f ( x )  x 
is increasing
x
and where it is decreasing.
Solution
 Thus, f is increasing on (– , –1) and (1, ), and decreasing
on (– 1, 0) and(0, 1):
y
4
2
–4
–2
2
–2
–4
Example 4, page 247
4
x
Relative Extrema
 The first derivative may be used to help us locate high
points and low points on the graph of f:
✦ High points are called relative maxima
✦ Low points are called relative minima.
 Both high and low points are called relative extrema.
y
Relative
Maxima
y = f(x)
Relative
Minima
x
Relative Extrema
Relative Maximum
 A function f has a relative maximum at x = c if
there exists an open interval (a, b) containing c
such that f(x)  f(c) for all x in (a, b).
Relative
Maxima
y
y = f(x)
x1
x2
x
Relative Extrema
Relative Minimum
 A function f has a relative minimum at x = c if
there exists an open interval (a, b) containing c
such that f(x)  f(c) for all x in (a, b).
y
y = f(x)
Relative
Minima
x3
x4
x
Finding Relative Extrema
 Suppose that f has a relative maximum at c.
 The slope of the tangent line to the graph must
change from positive to negative as x increases.
 Therefore, the tangent line to the graph of f at point
(c, f(c)) must be horizontal, so that f ′(x) = 0 or f ′(x) is
undefined.
y
f ′(x) > 0
f ′(x) = 0
f ′(x) < 0
a
c
b
x
Finding Relative Extrema
 Suppose that f has a relative minimum at c.
 The slope of the tangent line to the graph must
change from negative to positive as x increases.
 Therefore, the tangent line to the graph of f at point
(c, f(c)) must be horizontal, so that f ′(x) = 0 or f ′(x) is
undefined.
y
f ′(x) > 0
f ′(x) < 0
f ′(x) = 0
a
c
b
x
Finding Relative Extrema
 In some cases a derivative does not exist for
particular values of x.
 Extrema may exist at such points, as the graph
below shows:
Relative
Maximum
y
Relative
Minimum
a
b
x
Critical Numbers
 We refer to a number in the domain of f that may give
rise to a relative extremum as a critical number.
Critical number of f
✦ A critical number of a function f is any number
x in the domain of f such that f ′(x) = 0 or f ′(x)
does not exist.
Critical Numbers




The graph below shows us several critical numbers.
At points a, b, and c, f ′(x) = 0.
There is a corner at point d, so f ′(x) does not exist there.
The tangent to the curve at point e is vertical, so f ′(x)
does not exist there either.
 Note that points a, b, and d are relative extrema, while
points c and e are not.
y
Relative
Extrema
Not Relative
Extrema
a
b
c
d
e
x
The First Derivative Test
 Procedure for Finding Relative Extrema of a Continuous
Function f
1. Determine the critical numbers of f.
2. Determine the sign of f ′(x) to the left and right of
each critical point.
a. If f ′(x) changes sign from positive to negative as
we move across a critical number c, then f(c) is a
relative maximum.
b. If f ′(x) changes sign from negative to positive as
we move across a critical number c, then f(c) is a
relative minimum.
c. If f ′(x) does not change sign as we move across a
critical number c, then f(c) is not a relative
extremum.
Examples
 Find the relative maxima and minima of f ( x )  x 2
Solution
 The derivative of f is f ′(x) = 2x.
 Setting f ′(x) = 0 yields x = 0 as
the only critical number of f.
 Since f ′(x) < 0 if x < 0
and
f ′(x) > 0 if x > 0
we see that f ′(x) changes sign
from negative to positive as
we move across 0.
 Thus, f(0) = 0 is a relative
minimum of f.
y
f(x) = x2
5
4
3
2
f ′(x) < 0
–2
f ′(x) > 0
1
–1
0
1
Relative
Minimum
Example 5, page 252
2
x
Examples
 Find the relative maxima and minima of f ( x )  x
2/3
Solution
–1/3
 The derivative of f is f ′(x) = 2/3x .
 f ′(x) is not defined at x = 0, is continuous everywhere else,
and is never equal to zero in its domain.
 Thus x = 0 is the only critical number of f.
 Since f ′(x) < 0 if x < 0
and
f ′(x) > 0 if x > 0
y
we see that f ′(x) changes
4
f(x) = x2/3
sign from negative to
positive as we move
2
across 0.
f´(x) > 0
f´(x) < 0
 Thus, f(0) = 0 is a
x
relative minimum of f.
–4
–2
2
4
Relative
Minimum
Example 6, page 252
Examples
 Find the relative maxima and minima of
f ( x )  x 3  3x 2  24 x  32
Solution
 The derivative of f and equate to zero:
f ( x )  3x 2  6 x  24  0
3( x 2  2 x  8)  0
3( x  4)( x  2)  0
 The zeros of f ′(x) are x = –2 and x = 4.
 f ′(x) is defined everywhere, so x = –2 and x = 4 are the only
critical numbers of f.
Example 7, page 253
Examples
 Find the relative maxima and minima of
f ( x )  x 3  3x 2  24 x  32
Solution
 Since f ′(x) > 0 if x < –2 and f ′(x) < 0 if 0 < x < 4,
we see that f ′(x) changes sign from positive to negative as
we move across –2.
Relative
 Thus, f(–2) = 60 is a
Maximum
y
60
f(x)
relative maximum.
40
20
–7 –5
–3
–1
–20
–40
–60
Example 7, page 253
1
3
5
7
x
Examples
 Find the relative maxima and minima of
f ( x )  x 3  3x 2  24 x  32
Solution
 Since f ′(x) < 0 if 0 < x < 4 and f ′(x) > 0 if, x > 4
we see that f ′(x) changes sign from negative to positive as
we move across 4.
Relative
 Thus, f(4) = – 48 is a
Maximum
y
60
f(x)
relative minimum.
40
20
–7 –5
–3
–1
–20
1
3
5
–40
–60
Example 7, page 253
Relative
Minimum
7
x
4.2
Applications of the Second Derivative
y
Relative
Maxima
y = f(x)
Relative
Minima
x
Concavity
 A curve is said to be concave upwards when the slope of
tangent line to the curve is increasing:
y
x
 Thus, if f is differentiable on an interval (a, b), then
f is concave upwards on (a, b) if f ′ is increasing on (a, b).
Concavity
 A curve is said to be concave downwards when the slope of
tangent line to the curve is decreasing:
y
x
 Thus, if f is differentiable on an interval (a, b), then
f is concave downwards on (a, b) if f ′ is decreasing on (a, b).
Theorem 2
 Recall that f ″(x) measures the rate of change of the slope
f ′(x) of the tangent line to the graph of f at the point (x, f(x)).
 Thus, we can use f ″(x) to determine the concavity of f.
a. If f ″(x) > 0 for each value of x in (a, b), then f is
concave upward on (a, b).
b. If f ″(x) < 0 for each value of x in (a, b), then f is
concave downward on (a, b).
Steps in Determining the Concavity of f
1. Determine the values of x for which f ″is zero or where
f ″ is not defined, and identify the open intervals
determined by these numbers.
2. Determine the sign of f ″ in each interval found in step 1.
To do this compute f ″(c), where c is any conveniently
chosen test number in the interval.
a. If f ″(c) > 0, f is concave upward on that interval.
b. If f ″(c) < 0, f is concave downward on that interval.
Examples
 Determine where the function f ( x )  x 3  3x 2  24 x  32
is concave upward and where it is concave downward.
Solution
2
 Here, f ( x )  3x  6 x  24 and f ( x )  6 x  6
 Setting f ″(c) = 0 we find
f ( x )  6 x  6  0
6( x  1)  0
which gives x = 1.
 So we consider the intervals (– , 1) and (1, ):
✦ f ″(x) < 0 when x < 1, so f is concave downward on (– , 1).
✦ f ″(x) > 0 when x > 1, so f is concave upward on (1, ).
Example 1, page 265
Examples
 Determine where the function f ( x )  x 3  3x 2  24 x  32
is concave upward and where it is concave downward.
Solution
 The graph confirms that f is concave downward on (– , 1)
and concave upward on (1, ):
60
y
f(x)
40
20
–7 –5
–3
–1
–20
–40
–60
Example 1, page 265
1
3
5
7
x
Examples
 Determine the intervals where the function
f ( x)  x 
is concave upward and concave downward.
Solution
2
1 and
 Here, f ( x )  1 

f ( x)  3
2
x
x
 So, f ″cannot be zero and is not defined at x = 0.
1
x
 So we consider the intervals (– , 0) and (0, ):
✦ f ″(x) < 0 when x < 0, so f is concave downward on (– , 0).
✦ f ″(x) > 0 when x > 0, so f is concave upward on (0, ).
Example 2, page 266
Examples
 Determine the intervals where the function
f ( x)  x 
1
x
is concave upward and concave downward.
Solution
 The graph confirms that f is concave downward on (– , 0)
and concave upward on (0, ):
y
4
2
–4
–2
2
–2
–4
Example 2, page 266
4
x
Inflection Point
 A point on the graph of a continuous function where the
tangent line exists and where the concavity changes is
called an inflection point.
 Examples:
y
Inflection
Point
x
Inflection Point
 A point on the graph of a continuous function where the
tangent line exists and where the concavity changes is
called an inflection point.
 Examples:
y
Inflection
Point
x
Inflection Point
 A point on the graph of a continuous function where the
tangent line exists and where the concavity changes is
called an inflection point.
 Examples:
y
Inflection
Point
x
Finding Inflection Points
To find inflection points:
1. Compute f ″(x).
2. Determine the numbers in the domain of f for
which f ″(x) = 0 or f ″(x) does not exist.
3. Determine the sign of f ″(x) to the left and right of
each number c found in step 2.
If there is a change in the sign of f ″(x) as we move
across x = c, then (c, f(c)) is an inflection point of f.
Examples
 Find the points of inflection of the function f ( x )  x 3
Solution
2
 We have f ( x )  3x and f ( x )  6 x
 So, f ″ is continuous everywhere and is zero for x = 0.
 We see that f ″(x) < 0 when x < 0, and f ″(x) > 0 when x > 0.
 Thus, we find that the graph of f :
✦ Has one and only inflection
y
f(x)
point at f(0) = 0.
4
✦ Is concave downward on
Inflection
the interval (– , 0).
Point 2
✦ Is concave upward on the
interval (0, ).
–2
2
–2
–4
Example 3, page 268
x
Examples
 Find the points of inflection of the function f ( x )   x  1
5/3
Solution
10
5
2/3


f
(
x
)

f
(
x
)

x

1
 We have
  and
1/3
3
9  x  1
 So, f ″ is not defined at x = 1, and f ″ is never equal to zero.
 We see that f ″(x) < 0 when x < 1, and f ″(x) > 0 when x > 1.
 Thus, we find that the graph of f :
✦ Has one and only
inflection point at f(1) = 0.
✦ Is concave downward on
the interval (– , 1).
✦ Is concave upward on the
interval (1, ).
y
3
2 Inflection
Point
1
–2
–1
–1
–2
–3
Example 4, page 268
1
2
f(x)
x
Applied Example: Effect of Advertising on Sales
 The total sales S (in thousands of dollars) of Arctic Air Co.,
which makes automobile air conditioners, is related to the
amount of money x (in thousands of dollars) the company
spends on advertising its product by the formula
S ( x )  0.01x 3  1.5x 2  200
 Find the inflection point of the function S.
 Discuss the meaning of this point.
Applied Example 7, page 270
(0  x  100)
Applied Example: Effect of Advertising on Sales
Solution
 The first two derivatives of S are given by
S ( x)  0.03x 2  3x
and
S ( x)  0.06 x  3
 Setting S ″(x) = 0 gives x = 50.
 So (50, S(50)) is the only candidate for an inflection point.
 Since S ″(x) > 0 for x < 50, and S ″(x) < 0 for x > 50, the point
(50, 2700) is, in fact, an inflection point of S.
 This means that the firm experiences diminishing returns
on advertising beyond $50,000:
✦ Every additional dollar spent on advertising increases
sales by less than previously spent dollars.
Applied Example 7, page 270
The Second Derivative Test
 Maxima occur when a curve is concave downwards, while
minima occur when a curve is concave upwards.
 This is the basis of the second derivative test:
1. Compute f ′(x) and f ″(x).
2. Find all the critical numbers of f at which f ′(x) = 0.
3. Compute f ″(c), if it exists, for each critical number c.
a. If f ″(c) < 0, then f has a relative maximum at c.
b. If f ″(c) > 0, then f has a relative minimum at c.
c. If f ″(c) = 0, then the test fails (it is inconclusive).
Example
 Determine the relative extrema of the function
f ( x )  x 3  3x 2  24 x  32
Solution
 We have
f ( x)  3x 2  6 x  24  3( x  4)( x  2)
so f ′(x) = 0 gives the critical numbers x = –2 and x = 4.
 Next, we have f ( x )  6 x  6
which we use to test the critical numbers:
f ( 2)  6( 2)  6  12  6  18  0
so, f(–2) = 60 is a relative maximum of f.
f (4)  6(4)  6  24  6  18  0
so, f(4) = –48 is a relative minimum of f.
Example 9, page 272
4.3
Curve Sketching
x = –2
y
x=2
x
y = –1
Vertical Asymptotes
 The line x = a is a vertical asymptote of the
graph of a function f if either
lim f ( x )   or  
or
x a
lim f ( x )   or  
x a 
Finding Vertical Asymptotes of Rational Functions
 Suppose f is a rational function
P( x )
f ( x) 
Q( x)
where P and Q are polynomial functions.
 Then, the line x = a is a vertical asymptote of
the graph of f if Q(a) = 0 but P(a) ≠ 0.
Example
 Find the vertical asymptotes of the graph of the function
x2
f ( x) 
4  x2
Solution
 f is a rational function with P(x) = x2 and Q(x) = 4 – x2.
 The zeros of Q are found by solving
4  x2  0
(2  x )(2  x )  0
giving x = –2 and x = 2.
 Examine x = –2:
P(–2) = (–2)2 = 4 ≠ 0, so x = –2 is a vertical asymptote.
 Examine x = 2:
P(2) = (2)2 = 4 ≠ 0, so x = 2 is also a vertical asymptote.
Example 1, page 285
Horizontal Asymptotes
 The line y = b is a horizontal asymptote of the
graph of a function f if either
lim f ( x)  b or
x 
lim f ( x)  b
x 
Example
 Find the horizontal asymptotes of the graph of the function
x2
f ( x) 
4  x2
Solution
 We compute
x2
1
1
lim
 lim

 1
x  4  x 2
x  4
0 1

1
x2
and so y = –1 is a horizontal asymptote.
 We compute
x2
1
1
lim
 lim

 1
x  4  x 2
x  4
0 1

1
x2
also yielding y = –1 as a horizontal asymptote.
Example 2, page 287
Example
 Find the horizontal asymptotes of the graph of the function
x2
f ( x) 
4  x2
Solution
 So, the graph of f has two vertical asymptotes x = –2
and x = 2, and one horizontal asymptote y = –1:
y
x = –2
x=2
x
y = –1
f(x)
Example 2, page 287
Asymptotes and Polynomials
 A polynomial function has no vertical asymptotes.
 To see this, note that a polynomial function P(x) can be
written as a rational function with a denominator
equal to 1.
P( x )
 Thus,
P( x ) 
1
 Since the denominator is never zero, P has no vertical
asymptotes.
Asymptotes and Polynomials
 A polynomial function has no horizontal asymptotes.
 If P(x) is a polynomial of degree greater or equal to 1,
then
lim P( x)
x 
and
lim P( x)
x 
are either infinity or minus infinity; that is, they do
not exist.
 Therefore, P has no horizontal asymptotes.
A Guide to Sketching a Curve
1. Determine the domain of f.
2. Find the x- and y-intercepts of f.
3. Determine the behavior of f for large absolute
values of x.
4. Find all horizontal and vertical asymptotes of f.
5. Determine the intervals where f is increasing and
where f is decreasing.
6. Find the relative extrema of f.
7. Determine the concavity of f.
8. Find the inflection points of f.
9. Plot a few additional points to help further identify
the shape of the graph of f and sketch the graph.
Examples
 Sketch the graph of the function f ( x )  x  6 x  9 x  2
3
2
Solution
1. The domain of f is the interval (– , ).
2. By setting x = 0, we find that the y-intercept is 2.
(The x-intercept is not readily obtainable)
3. Since
lim  x 3  6 x 2  9 x  2   
x 
lim  x 3  6 x 2  9 x  2   
x 
we see that f decreases without bound as x decreases
without bound and that f increases without bound when
x increases without bound.
4. Since f is a polynomial function, there are no asymptotes.
Example 3, page 288
Examples
 Sketch the graph of the function f ( x )  x  6 x  9 x  2
3
2
Solution
5. f ( x)  3x 2  12 x  9  3( x  3)( x  1)
Setting f ′(x) = 0 gives x = 1 and x = 3 as critical points.
Testing with different values of x we find that f ′(x) > 0
when x < 1, f ′(x) < 0 when 1 < x < 3, and f ′(x) > 0 when x > 3.
Thus, f is increasing in the intervals (– , 1) and (3, ),
and f is decreasing in the interval (1, 3).
Example 3, page 288
Examples
 Sketch the graph of the function f ( x )  x  6 x  9 x  2
3
2
Solution
6. f ′ changes from positive to negative as we go across x = 1,
so a relative maximum of f occurs at (1, f(1)) = (1, 6).
f ′ changes from negative to positive as we go across x = 3,
so a relative minimum of f occurs at (3, f(3)) = (1, 2).
Example 3, page 288
Examples
 Sketch the graph of the function f ( x )  x  6 x  9 x  2
3
2
Solution
7. f ( x )  6 x  12  6( x  2)
which is equal to zero when x = 2.
Testing with different values of x we find that f ″(x) < 0
when x < 2 and f ″(x) > 0 when 2 < x.
Thus, f is concave downward in the interval (– , 2) and
concave upward in the interval (2, ).
8. Since f ″(2) = 0, we have an inflection point at (2, f(2)) = (2, 4).
Example 3, page 288
Examples
 Sketch the graph of the function f ( x )  x  6 x  9 x  2
3
2
Solution
 Summarizing, we’ve found the following:
✦ Domain: (– , ).
✦ Intercept: (0, 2).
✦ lim f ( x )   and lim f ( x )  
x 
x
✦ Asymptotes: None.
✦ f is increasing in the intervals (– , 1) and (3, ), and
✦
✦
✦
✦
f is decreasing in the interval (1, 3).
A relative maximum of f occurs at (1, 6).
A relative minimum of f occurs at (1, 2).
f is concave downward in the interval (– , 2) and
f is concave upward in the interval (2, ).
f has an inflection point at (2, 4).
Example 3, page 288
Examples
 Sketch the graph of the function f ( x )  x  6 x  9 x  2
3
2
Solution
 Sketch the graph:
y
7
(1, 6)
6
5
Relative maximum
(2, 4)
4
3
Inflection point
(3, 2)
2
1
Intercept
1
Example 3, page 288
Relative minimum
2
3
4
5
6
x
Examples
 Sketch the graph of the function f ( x )  x  6 x  9 x  2
3
2
Solution
 Sketch the graph:
y
7
6
5
4
3
2
1
1
Example 3, page 288
2
3
4
5
6
x
Examples
x 1
 Sketch the graph of the function f ( x ) 
x 1
Solution
1. f is undefined when x = 1, so the domain of f is the set of
all real numbers other than x = 1.
2. Setting y = 0, gives an x-intercept of –1.
Setting x = 0, gives an y-intercept of –1.
Example 4, page 290
Examples
x 1
 Sketch the graph of the function f ( x ) 
x 1
Solution
x 1
x 1
 1 and lim
1
3. Since lim
x  x  1
x  x  1
we see that f(x) approaches the line y = 1 as |x| becomes
arbitrarily large.
✦ For x > 1, f(x) > 1, so f approaches the line y = 1 from above.
✦ For x < 1, f(x) < 1, so f approaches the line y = 1 from below.
4. From step three we conclude that y = 1 is a horizontal
asymptote of f.
Also, the straight line x = 1 is a vertical asymptote of f.
Example 4, page 290
Examples
x 1
 Sketch the graph of the function f ( x ) 
x 1
Solution
5.
( x  1)(1)  ( x  1)(1)
2

f ( x) 

2
( x  1)
( x  1)2
So, f ′(x) is discontinuous at x = 1 and is never equal to zero.
Testing we find that f ′(x) < 0 wherever it is defined.
6. From step 5 we see that there are no critical numbers of f,
since f ′(x) is never equal to zero.
Example 4, page 290
Examples
x 1
 Sketch the graph of the function f ( x ) 
x 1
Solution
d
4
7.
2
3



f ( x) 
2( x  1)   4( x  1) 

dx
( x  1)3
Testing with different values of x we find that f ″(x) < 0
when x < 1 and f ″(x) > 0 when 1 < x.
Thus, f is concave downward in the interval (– , 1) and
concave upward in the interval (1, ).
8. From point 7 we see there are no inflection points of f, since
f ″(x) is never equal to zero.
Example 4, page 290
Examples
x 1
 Sketch the graph of the function f ( x ) 
x 1
Solution
 Summarizing, we’ve found the following:
✦ Domain: (– , 1) U (1, ).
✦ Intercept: (0, –1); (–1, 0).
✦
lim f ( x )  1 and lim f ( x )  1
x 
✦ Asymptotes:
✦
✦
✦
✦
x 
x = 1 is a vertical asymptote.
y = 1 is a horizontal asymptote.
f is decreasing everywhere in the domain of f.
Relative extrema: None.
f is concave downward in the interval (– , 1) and
f is concave upward in the interval (1, ).
f has no inflection points.
Example 4, page 290
Examples
x 1
 Sketch the graph of the function f ( x ) 
x 1
Solution
 Sketch the graph:
y
4
f ( x) 
2
x 1
x 1
y=1
–2
2
–2
Example 4, page 290
x=1
4
x
4.4
Optimization I
y
Absolute
maximum
Relative
maximum
Absolute
minimum
Relative
minimum
a
x1
x2
x3
b
x
Absolute Extrema
 The absolute extrema of a function f
✦ If f(x)  f(c) for all x in the domain of f, then f(c)
is called the absolute maximum value of f.
✦ If f(x)  f(c) for all x in the domain of f, then f(c)
is called the absolute minimum value of f.
Examples
 f has an absolute minimum at (0, 0):
y
f ( x)  x 2
4
2
–2
–1
1
Absolute minimum
2
x
Examples
 f has an absolute maximum at (0, 4):
y
Absolute maximum
4
f ( x)  4  x 2
2
–2
–1
1
2
x
Examples
 f has an absolute minimum at (  2 / 2, 1 / 2):
and an absolute maximum at ( 2 / 2,1 / 2):
1
y
Absolute maximum
1/2
f ( x)  x 1  x 2
x
–1
1
–1/2
Absolute minimum
–1
Examples
 f has no absolute extrema:
y
7
f ( x)  x3  6 x 2  9 x  2
6
5
4
3
2
1
1
2
3
4
5
6
x
Theorem 3
Absolute Extrema in a Closed Interval
 If a function f is continuous on a closed interval
[a, b], then f has both an absolute maximum
value and an absolute minimum value on [a, b].
Example
 The relative minimum of f at x3 is also the absolute
minimum of f.
 The right endpoint b of the interval [a, b] gives rise to the
absolute maximum value f(b) of f.
y
Relative
maximum
Absolute
maximum
Relative
maximum
Absolute
minimum
Relative
minimum
a
x1
x2
x3
b
x
Finding Absolute Extrema
To find the absolute extrema of f on a closed interval [a, b].
1. Find the critical numbers of f that lie on (a, b).
2. Compute the value of f at each critical number found in
step 1 and compute f(a) and f(b).
3. The absolute maximum value and absolute minimum
value of f will correspond to the largest and smallest
numbers, respectively, found in step 2.
Examples
 Find the absolute extrema of the function F(x) = x2 defined
on the interval [–1, 2].
Solution
 The function F is continuous on the closed interval [–1, 2]
and differentiable on the open interval (–1, 2).
 Setting F ′ = 0, we get F ′(x) = 2x = 0, so there is only one
critical point at x = 0.
 So, F(–1) = (–1)2 = 1, F(0) = (0)2 = 0, and F(2) = (2)2 = 4.
 It follows that 0 is the absolute minimum of F, and 4 is the
absolute maximum of F.
Example 1, page 300
Examples
 Find the absolute extrema of the function F(x) = x2 defined
on the interval [–1, 2].
Solution
y
4
Absolute
maximum
3
2
f ( x)  x 2
1
–2
Example 1, page 300
–1
x
1
Absolute
minimum
2
Examples
 Find the absolute extrema of the function
f ( x)  x 3  2 x 2  4 x  4
defined on the interval [0, 3].
Solution
 The function f is continuous on the closed interval [0, 3]
and differentiable on the open interval (0, 3).
 Setting f ′ = 0, we get
f ( x)  3x 2  4 x  4  (3x  2)( x  2)  0
which gives two critical points at x = – 2/3 and x = 2.
 We drop x = – 2/3 since it lies outside the interval [0, 3].
 So, f(0) = 4, f(2) = – 4, and f(3) = 1.
 It follows that – 4 is the absolute minimum of f, and 4 is
the absolute maximum of f.
Example 2, page 300
Examples
 Find the absolute extrema of the function
f ( x)  x 3  2 x 2  4 x  4
defined on the interval [0, 3].
Solution
y
f ( x)  x 3  2 x 2  4 x  4
4
2
Absolute
maximum
x
–2
2
4
–2
–4
Example 2, page 300
Absolute
minimum
Applied Example: Maximizing Profits
 Acrosonic’s total profit (in dollars) from manufacturing
and selling x units of their model F speakers is given by
P( x )  0.02 x 2  300 x  200,000
(0  x  20,000)
 How many units of the loudspeaker system must Acrosonic
produce to maximize profits?
Solution
 To find the absolute maximum of P on [0, 20,000], first find
the stationary points of P on the interval (0, 20,000).
 Setting f ′ = 0, we get
P( x )  0.04 x  300  0
300
x
 7500
0.04
which gives us only one stationary point at x = 7500.
Applied Example 4, page 301
Applied Example: Maximizing Profits
 Acrosonic’s total profit (in dollars) from manufacturing
and selling x units of their model F speakers is given by
P( x )  0.02 x 2  300 x  200,000
(0  x  20,000)
 How many units of the loudspeaker system must Acrosonic
produce to maximize profits?
Solution
 Evaluating the only stationary point we get
P(7500) = 925,000
 Evaluating the endpoints we get
P(0) = –200,000
P(20,000) = –2,200,000
 Thus, Acrosonic will realize the maximum profit of
$925,000 by producing 7500 speakers.
Applied Example 4, page 301
Applied Example: Maximizing Profits
 Acrosonic’s total profit (in dollars) from manufacturing
and selling x units of their model F speakers is given by
P( x )  0.02 x 2  300 x  200,000
(0  x  20,000)
 How many units of the loudspeaker system must Acrosonic
(Thousands of dollars)
produce to maximize profits?
Solution
y
1000
800
Maximum
Profit
600
P( x )
400
200
2
–200
Applied Example 4, page 301
4
6
8 10 12 14
(Thousands of speakers)
16
x
4.5
Inventory Level
Optimization II
x
x
2
Average
inventory
Time
Guidelines for Solving Optimization Problems
1. Assign a letter to each variable mentioned in the problem.
If appropriate, draw and label a figure.
2. Find an expression for the quantity to be optimized.
3. Use the conditions given in the problem to write the
quantity to be optimized as a function f of one variable.
Note any restrictions to be placed on the domain of f from
physical considerations of the problem.
4. Optimize the function f over its domain using the methods
of Section 4.4.
Applied Maximization Problem: Packaging
 By cutting away identical squares from each corner of a
rectangular piece of cardboard and folding up the resulting
flaps, the cardboard may be turned into an open box.
 If the cardboard is 16 inches long and 10 inches wide, find the
dimensions of the box that will yield the maximum volume.
10
16
Applied Example 2, page 314
Applied Maximization Problem: Packaging
Solution
1. Let x denote the length in inches of one side of each of the
identical squares to be cut out of the cardboard.
The dimensions of the box are (16 – 2x) by (10 – 2x) by x in.
x
10 10 – 2x
x
x
16 – 2x
16
Applied Example 2, page 314
x
Applied Maximization Problem: Packaging
Solution
2. Let V denote the volume (in cubic inches) of the resulting
box. The volume,
V  (16  2 x)(10  2 x) x
 4( x3  13x 2  40 x)
is the quantity to be maximized.
x
10 – 2x
Applied Example 2, page 314
16 – 2x
Applied Maximization Problem: Packaging
Solution
3. Each side of the box must be nonnegative, so x must satisfy
the inequalities x  0, 16 – 2x  0, and 10 – 2x  0.
✦ All these inequalities are satisfied if 0  x  5.
✦ Therefore, the problem at hand is equivalent to finding
the absolute maximum of
V  f ( x)  4( x 3  13x 2  40 x)
on the closed interval [0, 5].
Applied Example 2, page 314
Applied Maximization Problem: Packaging
Solution
4. f is continuous on [0, 5]. Setting f ′(x) = 0 we get
f ( x )  4(3x 2  26 x  40)
 4(3x  20)( x  2)
which yields the critical numbers x = 20/3 and x = 2.
We discard x = 20/3 for being outside the interval [0, 5].
We evaluate f at the critical point and at the endpoints:
f(0) = 0
f(2) = 144
f(5) = 0
Thus, the volume of the box is maximized by taking x = 2.
The resulting dimensions of the box are 12″ ☓ 6″ ☓ 2″.
Applied Example 2, page 314
Applied Minimization Problem: Inventory Control
 Dixie’s Import-Export is the sole seller of the Excalibur




250 cc motorcycle.
Management estimates that the demand for these
motorcycles will be 10,000 for the coming year and that
they will sell at a uniform rate throughout the year.
The cost incurred in ordering each shipment of
motorcycles is $10,000, and the cost per year of storing
each motorcycle is $200.
Dixie’s management faces the following problem:
✦ Ordering too many motorcycles at one time increases
storage cost.
✦ On the other hand, ordering too frequently increases the
ordering costs.
How large should each order be, and how often should
orders be placed, to minimize ordering and storage costs?
Applied Example 5, page 317
Applied Minimization Problem: Inventory Control
Inventory Level
Solution
 Let x denote the number of motorcycles in each order.
 Assuming each shipment arrives just as the previous
shipment is sold out, the average number of motorcycles in
storage during the year is x/2, as shown below:
 Thus, Dixie’s storage cost for the year is given by 200(x/2),
or 100x dollars.
x
x
2
Average
inventory
Time
Applied Example 5, page 317
Applied Minimization Problem: Inventory Control
Solution
 Next, since the company requires 10,000 motorcycles for the
year and since each order is for x motorcycles, the number
of orders required is
10, 000
x
 This gives an ordering cost of
 10,000  100,000,000
10,000 

x
 x 
dollars for the year.
 Thus, the total yearly cost incurred by Dixie’s, including
ordering and storage costs, is given by
100,000,000
C ( x )  100 x 
x
Applied Example 5, page 317
Applied Minimization Problem: Inventory Control
Solution
 The problem is reduced to finding the absolute minimum of
the function C in the interval (0,10,000].
 To accomplish this, we compute
100,000,000

C ( x )  100 
x2
 Setting C′(x) =0 and solving we obtain x = +1000.
 We reject the negative for being outside the domain.
 So we have x = 1000 as the only critical number.
Applied Example 5, page 317
Applied Minimization Problem: Inventory Control
Solution
 So we have x = 1000 as the only critical number.
 Now we find
200,000,000

C ( x) 
x3
 Since C″(1000) > 0, the second derivative test implies that
x = 1000 is a relative minimum of C.
 Also, since C″(1000) > 0 for all x in (0,10,000], the function
C is concave upward everywhere so that also gives the
absolute minimum of C.
 Thus, to minimize the ordering and storage costs, Dixie’s
should place 10,000/1000, or 10, orders per year, each for a
shipment of 1000 motorcycles.
Applied Example 5, page 317
End of
Chapter