Download Section 3.1 (cont.)

Lesson 26
Definition
First
Derivative
Test
Examples
Section 3.1 (cont.)
March 14th, 2014
Lesson 26
Definition
First
Derivative
Test
Examples
In this lesson we will discuss relative extrema. We first give a
definition for relative minima and relative maxima, and then we
present a test for determining the relative extrema of a
differentiable function.
Lesson 26
Definition
Definition
First
Derivative
Test
The graph of a function f (x) has a relative maximum at
x = c if f (c) ≥ f (x) for all x in an interval a < x < b
containing c. It has a relative minimum at x = c if
f (c) ≤ f (x) for all x in an interval a < x < b containing c.
Relative maxima and relative minima are called relative
extrema.
Examples
To visualize a relative minimum or maximum, picture a
parabola. If it opens up, its vertex will be a minimum. If it
opens down, its vertex will be a maximum.
Lesson 26
Definition
First
Derivative
Test
Examples
We must give one more definition before we discuss how to
determine the relative extrema of a function.
Definition
A number c in the domain of f (x) is called a critical number
if f 0 (c) = 0 or f 0 (c) does not exist. The corresponding point
(c, f (c)) is called a critical point.
A key part of the definition is that a critical number must be in
the domain of the given function. If f 0 (c) is undefined but c is
NOT in the domain of f , then c is NOT a critical number of f .
Lesson 26
Definition
First
Derivative
Test
The First Derivative Test for Relative Extrema
Let c be a critical number for f (x). The critical point
P(c, f (c)) is
a relative maximum if f 0 (x) > 0 to the left of c and
f 0 (x) < 0 to the right of c.
Examples
f0 > 0
f0 < 0
a relative minimum if f 0 (x) < 0 to the left of c and
f 0 (x) > 0 to the right of c.
f0 < 0
f0 > 0
not a relative extremum if f 0 (x) has the same sign on both
sides of c.
f0 > 0
f0 > 0
f0 < 0
f0 < 0
Lesson 26
Note: Not every critical point is a relative extremum.
Example
Definition
First
Derivative
Test
Examples
Determine the critical numbers of the given function and
classify each critical point as a relative maximum, a relative
minimum, or neither.
f (x) = 324x − 72x 2 + 4x 3
First we find the critical numbers of f (x):
f 0 (x) = 324 − 144x + 12x 2
12x 2 − 144x + 324 = 0 ⇒ x 2 − 12x + 27 = 0
⇒ (x − 3)(x − 9) = 0
Thus the critical numbers of f (x) are x = 3, 9. Note that
these are in the domain of f .
Lesson 26
Definition
First
Derivative
Test
Since the first derivative comes down to computing the sign of
the derivative, we make a sign chart (just like in the last
lesson).
3
Examples
9
f 0 (x) = 12(x − 3)(x − 9)
f 0 (0) = 12(−3)(−9) > 0, f 0 (4) = 12(1)(−4) < 0,
f 0 (10) = 12(7)(1) > 0
We can now decorate the number line. Since f 0 is positive if
and only if f is increasing, we can use either + signs or upward
arrows (similarly for − signs or downward arrows).
Lesson 26
Definition
First
Derivative
Test
3
9
Examples
Looking at this chart, it is clear that f (x) has a relative
maximum at x = 3 and a relative minimum at x = 9. Since
f (3) = 432 and f (9) = 0, we have:
relative min. : (9, 0),
relative max. : (3, 432)
Lesson 26
Definition
F (x) =
x2
x−1
(x−1)2x−x 2
(x−1)2
2x 2 −2x−x 2
(x−1)2
x 2 −2x
(x−1)2
First
Derivative
Test
F 0 (x) =
Examples
F 0 (x) = 0 : x(x − 2) = 0 ⇒ x = 0, 2
=
=
=
x(x−2)
(x−1)2
F 0 (x) DNE : (x − 1)2 = 0 ⇒ x − 1 = 0 ⇒ x = 1
Since 1 is not in the domain of F (x), x = 1 is NOT a
critical number. Thus the critical numbers of F (x) are
x = 0, 2. However, the sign of F 0 (x) can still change
around the value x = 1, and so we should still include it in
our number line.
Lesson 26
0
Definition
First
Derivative
Test
Examples
F 0 (−1) =
F 0 ( 32 ) =
−1(−1−2)
(−1−1)2
3 3
( −2)
2 2
( 32 −1)2
1
F 0 ( 12 ) =
> 0,
< 0,
2
F 0 (3) =
1 1
( −2)
2 2
( 21 −1)2
3(3−2)
(3−2)2
<0
>0
So F has a relative maximum at x = 0 and a relative minimum
at x = 2. Thus we have
relative max. : (0, 0)
relative min. : (2, 4)
Note that the relative minimum is higher than the relative
maximum. This is alright, because F (x) has a discontinuity at
x = 1.
Lesson 26
Example
Definition
First
Derivative
Test
Examples
Let p = (10 − 3x)2 for 0 ≤ x ≤ 3 be the price at which x
hundred units of a certain commodity will be sold, and let
R(x) = xp(x) be the revenue obtained from the sale of the x
units. For what level of production is revenue maximized?
The question asks us to find the level of production (between 0
and 3) for which revenue is maximized. Thus we should find
the maximum of R(x) on the interval 0 ≤ x ≤ 3.
R(x) = x(10 − 3x)2 ⇒ R 0 (x) = (10 − 3x)2 + 2x(10 − 3x)(−3)
= (10 − 3x)(10 − 9x)
So R 0 (x) = 0 at x =
10 10
9 , 3 .
Since
10
3
> 3, we throw it out.
Lesson 26
Thus x =
10
9
is the only critical number of R(x).
Definition
First
Derivative
Test
0
Examples
R 0 (1) = 7(1) > 0,
10
9
3
R 0 (2) = 4(−8) < 0
Thus R(x) has a relative maximum at x = 10
9 . Since R(x)
increases as it moves towards 10
and
decreases
as it moves
9
10
away from 10
,
R(
)
is
the
(absolute)
maximum
of R(x) on
9
9
0 ≤ x ≤ 3. Therefore revenue is maximized when x = 10
9
hundred units are produced.
Lesson 26
Example
Definition
First
Derivative
Test
Examples
Commissioners of a certain city determine that when x million
dollars are spent on controlling pollution, the percentage of
pollution removed is given by
√
100 x
P(x) =
0.04x 2 + 12
What expenditure results in the largest percentage of pollution
removal?
We must find the maximum of P(x) (since P(x) measures the
percentage of removal of pollution and we want to find the last
percentage of removal).
Lesson 26
P 0 (x) =
Definition
First
Derivative
Test
=
Examples
P 0 (x) = 0
(0.04x 2 + 12)50x −1/2 − 100x 1/2 (0.08x)
(0.04x 2 + 12)2
2x 3/2 + 600x −1/2 − 8x 3/2
(0.04x 2 + 12)2
=√
6(100 − x 2 )
x(0.04x 2 + 12)2
⇒
100 − x 2 = 0
⇒ x = −10, 10
√
is undefined at the zeros of x and 0.04x 2 + 12. Since
0.04x 2 + 12 is the denominator of P(x), its zeros are not in the
domain of P(x) and so can’t be critical numbers. However
x = 0 is a critical number.
P 0 (x)
Lesson 26
Definition
First
Derivative
Test
So the critical numbers of P(x) are x = −10, 0, 10. Since x is
measuring millions of dollars spent controlling pollution, we
throw out all negative values of x.
Examples
0
P 0 (1) ≈ 4.098 > 0
10
P 0 (11) ≈ −0.134 < 0
Thus x = 10 is the relative (and so absolute) maximum of
P(x) for x ≥ 0. Therefore an expenditure of $10,000,000
results in the largest percentage of pollution removal.