Lesson 27
Concave up and Concave down
October 23, 2013
Lesson 27 Concave up and Concave down
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Review (Lesson 5 – Section 1.4)
A box with a square base has a surface area of 600 in2 . Write the volume
of the box as a function of the width of the base.
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Review (Lesson 5 – Section 1.4)
A box with a square base has a surface area of 600 in2 . Write the volume
of the box as a function of the width of the base.
Call the width of the base x, and the height of the box h. Then the
volume is
V = x 2 h,
but to get a function of x, we need to replace h as a function of x. The
surface area of the box is A = 2x 2 + 4xh = 600, so
h=
and
V =x
2
600 − 2x 2
,
4x
600 − 2x 2
4x
1
= 150x − x 3 .
2
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Concavity
Lesson 27 is about interpreting the meaning of the second derivative.
Remember the procedure for the first derivative test:
1. Find critical points: where f 0 = 0 or f 0 DNE.
2. Test points to find where f 0 > 0 or f 0 < 0.
3. For each critical point, determine if it is a max, a min, or neither.
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Concavity
Lesson 27 is about interpreting the meaning of the second derivative.
Remember the procedure for the first derivative test:
1. Find critical points: where f 0 = 0 or f 0 DNE.
2. Test points to find where f 0 > 0 or f 0 < 0.
3. For each critical point, determine if it is a max, a min, or neither.
The points where f 00 = 0 or f 00 DNE are possible inflection points.
If f 00 > 0, the function is concave up;
if f 00 < 0, the function is concave down.
The inflection points are those where the concavity changes.
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Concave up
One function that is always concave up is y = x 2 .
All of these thing are the same:
1. The function f is concave up.
2. The derivative f 0 is increasing.
3. The second derivative f 00 is positive.
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Concave down
One function that is always concave down is y = −x 2 , for which y 00 = −2.
All three of these things are the same:
1. The function f is concave down.
2. The derivative f 0 is decreasing.
3. The second derivative f 00 is negative.
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Inflection point
The simplest function with an inflection point is y = x 3 , with y 00 = 6x.
The previous two examples did not have inflection points, because their
second derivative was either always positive or always negative.
Since x 3 is concave down on (−∞, 0) and concave up on (0, ∞), it has an
inflection point at (0, 0).
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Example 1 (HW #1–6)
Find where f (x) = 6x 4 − 13 x 3 + 19x − 1 is concave up and concave down.
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Example 1 (HW #1–6)
Find where f (x) = 6x 4 − 13 x 3 + 19x − 1 is concave up and concave down.
The second derivative is 72x 2 − 2x = 2x(36x − 1).
Therefore the possible inflection points are at x = 0,
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1
.
36
Example 1 (HW #1–6)
Find where f (x) = 6x 4 − 13 x 3 + 19x − 1 is concave up and concave down.
The second derivative is 72x 2 − 2x = 2x(36x − 1).
Therefore the possible inflection points are at x = 0,
1
.
36
Testing points, we have:
(−∞, 0): f 00 (−1) = 74 > 0, concave up.
1
1
00
(0, 1/36): f (1/72) =
−
< 0, concave down.
36
2
(1/36, ∞): f 00 (1) = 70 > 0, concave up.
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Example 2
Find the inflection points of f (x) =
√
3
x − 4.
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Example 2
Find the inflection points of f (x) =
√
3
x − 4.
2
The second derivative is − (x − 4)−5/3 . This does not exist when x = 4,
9
so that is the only possible inflection point.
(−∞, 4): f 00 (3) = 2/9 > 0, so concave up.
(4, ∞): f 00 (5) = −2/9 < 0, so concave down.
Therefore (4, 0) is an inflection point.
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Example 2
Find the inflection points of f (x) =
√
3
x − 4.
2
The second derivative is − (x − 4)−5/3 . This does not exist when x = 4,
9
so that is the only possible inflection point.
(−∞, 4): f 00 (3) = 2/9 > 0, so concave up.
(4, ∞): f 00 (5) = −2/9 < 0, so concave down.
Therefore (4, 0) is an inflection point.
Warning: (x − 4)−1/3 is not defined at x = 4.
√
x − 4 is not defined for x ≤ 4.
Neither function has inflection points.
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Second derivative test
Sometimes it is easier to calculate the second derivative than it is to
evaluate the first derivative many times. In these cases, you can use the
concavity of a function to decide whether a critical point is maximum,
minimum, or neither.
If f 00 > 0 at a critical point, then it is a minimum;
If f 00 < 0 at a critical point, then it is a maximum;
Otherwise, one needs to use the first derivative test.
Warning! This is the opposite of what you might expect!
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Example 3 (HW #7–11)
Find the minimums and maximums of f (x) = x 2 (x − 1)2 .
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Example 3 (HW #7–11)
Find the minimums and maximums of f (x) = x 2 (x − 1)2 .
We have f 0 (x) = 2x(x − 1)2 + 2x 2 (x − 1) = 2x(x − 1)(2x − 1).
The critical points are at x = 0, 1/2, 1.
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Example 3 (HW #7–11)
Find the minimums and maximums of f (x) = x 2 (x − 1)2 .
We have f 0 (x) = 2x(x − 1)2 + 2x 2 (x − 1) = 2x(x − 1)(2x − 1).
The critical points are at x = 0, 1/2, 1.
Now we test each with the second derivative, which is:
f 00 (x) = 2(x − 1)(2x − 1) + 2x(2x − 1) + 4x(x − 1).
So we calculate,
x = 0: f 00 (0) = 2 + 0 + 0 > 0, so x = 0 is minimum.
x = 1/2: f 00 (1/2) = 0 + 0 + −1 < 0, so x = 1/2 is maximum.
x = 1: f 00 (1) = 0 + 2 + 0 > 0, so x = 1 is minimum.
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Example 4
Find the minimums and maximums of f (x) =
Lesson 27 Concave up and Concave down
x
x −1
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2
.
Example 4
Find the minimums and maximums of f (x) =
The second derivative is −2
2
(x − 1) − x
(x − 1)2
(x − 1)3 − 3x(x − 1)2
(x − 1)6
x
x −1
There are critical points at x = 0, 1.
The first derivative is f 0 = 2
x
x −1
.
=−
=
2x
.
(x − 1)3
4x + 2
(x − 1)4
x = 0: f 00 (0) = 6 > 0, minimum.
x = 1: f 00 (1) DNE, inconclusive.
Notice that f (1) DNE, so this critical point is neither a minimum or
maximum. If it were defined, we would have to use the first derivative test.
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Review
Today:
1. Used the second derivative to find intervals of concave up and down.
2. Found inflection points by checking where the concavity changes.
3. Found minimums and maximums with the second derivative test.
Friday:
1. Finish Homework 27.
2. Read Example 3.2.6 on p. 226.
3. Be ready for a quiz on concavity and the second derivative test.
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