The Mean Value Theorem gives us tests for determining the shape of curves between critical points.

1. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Sec on 4.2
Deriva ves and the Shapes of Curves
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
April 11, 2011
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Notes
Announcements
Quiz 4 on Sec ons 3.3,
3.4, 3.5, and 3.7 this
week (April 14/15)
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
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Notes
Objectives
Use the deriva ve of a func on
to determine the intervals along
which the func on is increasing
or decreasing (The
Increasing/Decreasing Test)
Use the First Deriva ve Test to
classify cri cal points of a
func on as local maxima, local
minima, or neither.
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2. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test
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Notes
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
c
Let f be con nuous on [a, b]
and differen able on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a) .
= f′ (c). a
b−a
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′ (c)(b − a)
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Why the MVT is the MITC Notes
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 con nuous on
[x, y] and differen able 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.
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3. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test
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.
Notes
Increasing Functions
Defini on
A func on f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.
An increasing func on “preserves order.”
I could be bounded or infinite, open, closed, or
half-open/half-closed.
Write your own defini on (muta s mutandis) of decreasing,
nonincreasing, nondecreasing
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Notes
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0
on an interval, then f is decreasing on that interval.
Proof.
It works the same as the last theorem. Assume f′ (x) > 0 on an
interval I. Pick two points x and y in I 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).
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4. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solu on
f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solu on
1
Since f′ (x) = is always posi ve, f(x) is always increasing.
. 1 + x2
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Notes
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
f′ (x) = 2x, which is posi ve when x > 0 and nega ve when x is.
We can draw a number line:
− 0 + f′
.
↘ 0 ↗ f
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Notes
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
− 0 + f′
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↘ 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, ∞)
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5. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
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Notes
The First Derivative Test
Theorem (The First Deriva ve Test)
Let f be con nuous on [a, b] and c a cri cal point of f in (a, b).
If f′ changes from posi ve to nega ve at c, then c is a local
maximum.
If f′ changes from nega ve to posi ve at c, then c is a local
minimum.
If f′ (x) has the same sign on either side of c, then c is not a local
extremum.
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Notes
Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test
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. 5
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6. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Concavity
Defini on
The graph of f is called concave upwards on an interval if it lies
above all its tangents on that interval. The graph of f is called
concave downwards on an interval if it lies below all its tangents on
that interval.
. .
concave up concave down
. We some mes say a concave up graph “holds water” and a concave
down graph “spills water”.
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Notes
Synonyms for concavity
Remark
“concave up” = “concave upwards” = “convex”
“concave down” = “concave downwards” = “concave”
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Notes
Inflection points mean change in concavity
Defini on
A point P on a curve y = f(x) is called an inflec on point if f is
con nuous at P and the curve changes from concave upward to
concave downward at P (or vice versa).
concave
up
inflec on point
.
concave
down
.
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7. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Testing for Concavity
Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval, then the graph of f is concave
upward on that interval.
If f′′ (x) < 0 for all x in an interval, then the graph of f is concave
downward on that interval.
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Notes
Testing for Concavity
Proof.
Suppose f′′ (x) > 0 on the interval I (which could be infinite). 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).
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Notes
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solu on
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
This is nega ve when x < −1/3, posi ve when x > −1/3, and 0
when x = −1/3
So f is concave down on the open interval (−∞, −1/3), concave
up on the open interval (−1/3, ∞), and has an inflec on point
at the point (−1/3, 2/27)
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8. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have
10 −1/3 4 −4/3
f′′ (x) = x − x
9 9
2 −4/3
= x (5x − 2)
9
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Notes
The Second Derivative Test
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous 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 of f.
If f′′ (c) > 0, then c is a local minimum of f.
Remarks
If f′′ (c) = 0, the second deriva ve test is inconclusive
We look for zeroes of f′ and plug them into f′′ to determine if
their f values are local extreme values.
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Notes
Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + + + f′′ = (f′ )′
sufficiently close to c. .
↗ c ↗ f′
Since f′′ = (f′ )′ , we know
0 f′
f′ is increasing near c.
c f
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9. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′ (c) = 0 and f′ is
increasing, f′ (x) < 0 for x + + + f′′ = (f′ )′
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close to c and less than c, c
↗ ↗ f′
and f′ (x) > 0 for x close
− 0 + f′
to c and more than c.
c f
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Notes
Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
This means f′ changes
sign from nega ve to + + + f′′ = (f′ )′
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posi ve at c, which ↗ c ↗ f′
means (by the First
− 0 + f′
Deriva ve Test) that f has
a local minimum at c. c
↘ min ↗ f
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Notes
Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solu on
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.
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10. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solu on
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Using the Second Derivative Test II Notes
Graph
Graph of f(x) = x2/3 (x + 2):
y
. x
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Notes
When the second derivative is zero
Remark
At inflec on points c, if f′ is differen able at c, then f′′ (c) = 0
If f′′ (c) = 0, must f have an inflec on point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have cri cal points at zero with a second deriva ve of
zero. But the first has a local min at 0, the second has a local max at
0, and the third has an inflec on point at 0. This is why we say 2DT
has nothing to say when f′′ (c) = 0.
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11. . V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
When first and second derivative are zero
func on deriva ves graph type
′ 3 ′
f (x) = 4x , f (0) = 0
f(x) = x4 . 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
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Notes
Summary
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: deriva ves can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing Test
and the Concavity Test
Techniques for finding extrema: the First Deriva ve Test and
the Second Deriva ve Test
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Notes
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