Lesson 20: Derivatives and the Shape of Curves (Section 041 handout)
1. Section 4.2
Derivatives and the Shapes of Curves
V63.0121.041, Calculus I
New York University
November 15, 2010
Announcements
Quiz this week in recitation on 3.3, 3.4, 3.5, 3.7
There is class on November 24
Announcements
Quiz this week in recitation
on 3.3, 3.4, 3.5, 3.7
There is class on November
24
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 2 / 31
Objectives
Use the derivative of a
function to determine the
intervals along which the
function is increasing or
decreasing (The
Increasing/Decreasing Test)
Use the First Derivative Test
to classify critical points of a
function as local maxima,
local minima, or neither.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 3 / 31
Notes
Notes
Notes
1
Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
2. Objectives
Use the second derivative of
a function to determine the
intervals along which the
graph of the function is
concave up or concave down
(The Concavity Test)
Use the first and second
derivative of a function to
classify critical points as
local maxima or local
minima, when applicable
(The Second Derivative
Test)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 4 / 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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 5 / 31
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
f (b) − f (a)
b − a
= f (c).
a
b
c
Another way to put this is that there exists a point c such that
f (b) = f (a) + f (c)(b − a)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 31
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Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
3. 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)(y − x)
So f (y) = f (x). Since this is true for all x and y in (a, b), then f is
constant.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 7 / 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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 8 / 31
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 31
Notes
Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
4. 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)(y − x) > 0.
So f (y) > f (x).
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 31
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
Since f (x) =
1
1 + x2
is always positive, f (x) is always increasing.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 31
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:
f
f
−
0
0 +
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 31
Notes
Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
5. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f (x) = x2/3
(x + 2).
Solution
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 31
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 14 / 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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 17 / 31
Notes
Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
6. Concavity
Definition
The graph of f is called concave up on an 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”.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 18 / 31
Inflection points indicate a change in concavity
Definition
A point P on a curve y = f (x) is called an inflection point if f is
continuous at P and the curve changes from concave upward to concave
downward at P (or vice versa).
concave
down
concave up
inflection point
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 19 / 31
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)
By MVT, there exists a c between a and x with
f (x) = f (a) + f (c)(x − a)
Since f is increasing, f (x) > L(x).
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 20 / 31
Notes
Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
7. Finding Intervals of Concavity I
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)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 31
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x2/3
(x + 2).
Solution
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 31
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 c is a local maximum.
If f (c) > 0, then 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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 31
Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
8. Proof of the Second Derivative Test
Proof.
Suppose f (c) = 0 and f (c) > 0. Since f is continuous, f (x) > 0 for
all x sufficiently close to c. Since f = (f ) , we know f is increasing near
c. Since f (c) = 0 and f is increasing, f (x) < 0 for x close to c and less
than c, and f (x) > 0 for x close to c and more than c. This means f
changes sign from negative to positive at c, which means (by the First
Derivative Test) that f has a local minimum at c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 31
Using the Second Derivative Test I
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 31
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x2/3
(x + 2)
Solution
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 31
Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
9. Using the Second Derivative Test II: Graph
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 31
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) = x4
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 28 / 31
When first and second derivative are zero
function derivatives graph type
f (x) = x4
f (x) = 4x3, f (0) = 0
min
f (x) = 12x2, f (0) = 0
g(x) = −x4
g (x) = −4x3, g (0) = 0
max
g (x) = −12x2, g (0) = 0
h(x) = x3
h (x) = 3x2, h (0) = 0
infl.
h (x) = 6x, h (0) = 0
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 31
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010
10. 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) = x4
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 30 / 31
Summary
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 31 / 31
Notes
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Section 4.2 : The Shapes of CurvesV63.0121.041, Calculus I November 15, 2010