The document is a lecture on derivatives and the shapes of curves. It covers the mean value theorem, testing for monotonicity using the first derivative test, and finding intervals of monotonicity. It also discusses concavity and the second derivative test. Examples are provided to demonstrate how to find the intervals where a function is increasing or decreasing using the first derivative test.
1. Section 4.2
Derivatives and the Shapes of Curves
V63.0121.027, Calculus I
November 12, 2009
Announcements
Final Exam Friday, December 18, 2:00–3:50pm
.
.
Image credit: cobalt123
. . . . . .
2. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
3. Recall: The Mean Value Theorem
Theorem (The Mean Value
Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a
. . . . . .
4. Recall: The Mean Value Theorem
Theorem (The Mean Value
Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a
. . . . . .
5. Recall: The Mean Value Theorem
Theorem (The Mean Value c
..
Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a
. . . . . .
6. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x < y. Then f is continuous
on [x, y] and differentiable on (x, y). By MVT there exists a point
z in (x, y) such that
f(y) − f(x)
= f′ (z) = 0.
y−x
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.
. . . . . .
7. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
9. What does it mean for a function to be increasing?
Definition
A function f is increasing on (a, b) if
f(x) < f(y)
whenever x and y are two points in (a, b) with x < y.
An increasing function “preserves order.”
Write your own definition (mutatis mutandis) of decreasing,
nonincreasing, nondecreasing
. . . . . .
10. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
then f is decreasing on (a, b).
. . . . . .
11. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
then f is decreasing on (a, b).
Proof.
It works the same as the last theorem. Pick two points x and y in
(a, b) with x < y. We must show f(x) < f(y). By MVT there exists
a point c in (x, y) such that
f(y) − f(x)
= f′ (c) > 0.
y−x
So
f(y) − f(x) = f′ (c)(y − x) > 0.
. . . . . .
13. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
. . . . . .
14. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
. . . . . .
15. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solution
f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
1
Since f′ (x) = is always positive, f(x) is always increasing.
1 + x2
. . . . . .
17. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
. . . . . .
18. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
0
.
. . . . . .
19. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
. . . . . .
20. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
. . . . . .
21. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing
on [0, ∞)
. . . . . .
23. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
. . . . . .
24. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
0
.. ×
.. .′ (x)
f
−
. 4/5 0
. f
.(x)
. . . . . .
25. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
.
+ 0 − ×
.. . . . .
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .
26. The First Derivative Test
Theorem (The First Derivative Test)
Let f be continuous on [a, b] and c a critical point of f in (a, b).
If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local
maximum.
If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local
minimum.
If f′ (x) has the same sign on (a, c) and (c, b), then c is not a
local extremum.
. . . . . .
27. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing
on [0, ∞)
. . . . . .
28. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
f′ (x) = 2x, which is positive when x > 0 and negative when x
is.
We can draw a number line:
−
. 0
.. .
+ .′
f
↘
. 0
. ↗
. f
.
m
. in
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing
on [0, ∞)
. . . . . .
29. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
.
+ 0 − ×
.. . . . .
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
. . . . . .
30. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solution
f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4)
3 3
The critical points are 0 and and −4/5.
−
. ×
.. .
+
. −1/3
x
0
.
−
. 0
.. .
+
.x+4
5
−
. 4/5
.
+ 0 − ×
.. . . . .
+ .′ (x)
f
↗
. − ↘ .
. 4/5 . 0 ↗
. f
.(x)
m
. ax . in
m
. . . . . .
31. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Derivative Test
Concavity
Definitions
Testing for Concavity
The Second Derivative Test
. . . . . .
32. Definition
The graph of f is called concave up on and interval I if it lies
above all its tangents on I. The graph of f is called concave down
on I if it lies below all its tangents on I.
. .
concave up concave down
We sometimes say a concave up graph “holds water” and a
concave down graph “spills water”.
. . . . . .
33. 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 at P (or vice versa).
.concave up
i
.nflection point
. .
.
concave
down
. . . . . .
34. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval I, then the graph of f is
concave upward on I
If f′′ (x) < 0 for all x in I, then the graph of f is concave
downward on I
. . . . . .
35. Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval I, then the graph of f is
concave upward on I
If f′′ (x) < 0 for all x in I, then the graph of f is concave
downward on I
Proof.
Suppose f′′ (x) > 0 on I. This means f′ is increasing on I. Let a and
x be in I. The tangent line through (a, f(a)) is the graph of
L(x) = f(a) + f′ (a)(x − a)
f(x) − f(a)
By MVT, there exists a c between a and x with = f′ (c).
x−a
So
f(x) = f(a) + f′ (c)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x)
. . . . . .
39. Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solution
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
This is negative when x < −1/3, positive when x > −1/3, and
0 when x = −1/3
So f is concave down on (−∞, −1/3), concave up on
(1/3, ∞), and has an inflection point at (−1/3, 2/27)
. . . . . .
43. Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) = x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor
. . . . . .
44. Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solution
10 −1/3 4 −4/3 2 −4/3
f′′ (x) = x − x = x (5x − 2)
9 9 9
The second derivative f′′ (x) is not defined at 0
Otherwise, x−4/3 is always positive, so the concavity is
determined by the 5x − 2 factor
So f is concave down on (−∞, 0], concave down on [0, 2/5),
concave up on (2/5, ∞), and has an inflection point when
x = 2/5
. . . . . .
45. The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in
(a, b) with f′ (c) = 0.
If f′′ (c) < 0, then f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.
. . . . . .
46. The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in
(a, b) with f′ (c) = 0.
If f′′ (c) < 0, then f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.
Remarks
If f′′ (c) = 0, the second derivative test is inconclusive (this
does not mean c is neither; we just don’t know yet).
We look for zeroes of f′ and plug them into f′′ to determine if
their f values are local extreme values.
. . . . . .
50. Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
. . . . . .
51. Example
Find the local extrema of f(x) = x3 + x2 .
Solution
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
Since f′′ (0) = 2 > 0, 0 is a local minimum.
. . . . . .
54. Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when
3
x = −4/5
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
. . . . . .
55. Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when
3
x = −4/5
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
So x = −4/5 is a local maximum.
. . . . . .
56. Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solution
1 −1/3
Remember f′ (x) = x (5x + 4) which is zero when
3
x = −4/5
10 −4/3
Remember f′′ (x) = x (5x − 2), which is negative when
9
x = −4/5
So x = −4/5 is a local maximum.
Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.
. . . . . .
58. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
. . . . . .
59. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
Consider these examples:
f (x ) = x 4 g(x) = −x4 h(x) = x3
. . . . . .
60. When first and second derivative are zero
function derivatives graph type
f′ (x) = 4x3 , f′ (0) = 0
f (x ) = x 4 min
f′′ (x) = 12x2 , f′′ (0) = 0 .
.
g′ (x) = −4x3 , g′ (0) = 0
g (x ) = −x4 max
g′′ (x) = −12x2 , g′′ (0) = 0
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x3 infl.
h′′ (x) = 6x, h′′ (0) = 0 .
. . . . . .
61. When the second derivative is zero
At inflection points c, if f′ is differentiable at c, then f′′ (c) = 0
Is it necessarily true, though?
Consider these examples:
f (x ) = x 4 g(x) = −x4 h(x) = x3
All of them have critical points at zero with a second derivative of
zero. But the first has a local min at 0, the second has a local
max at 0, and the third has an inflection point at 0. This is why
we say 2DT has nothing to say when f′′ (c) = 0.
. . . . . .
62. What have we learned today?
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: derivatives can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing
Test and the Concavity Test
Techniques for finding extrema: the First Derivative Test and
the Second Derivative Test
. . . . . .
63. What have we learned today?
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: derivatives can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing
Test and the Concavity Test
Techniques for finding extrema: the First Derivative Test and
the Second Derivative Test
Next week: Graphing functions!
. . . . . .