3 4 Concavity 2nd Derv Test
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Transcript 3 4 Concavity 2nd Derv Test
3-4: Concavity and the Second Derivative Test
Objectives:
Assignment:
1. To determine the
concavity and points
of inflection of the
graph of a function
β’ P. 195-197: 1-25 odd,
49-52, 76, 79-82
2. To apply the Second
Derivative Test to
find relative extrema
β’ P. 195: 17-39 odd
Concave Upward
Informally, a
graph is
concave
upward if it
curves up.
Concave Downward
Informally,
a graph is
concave
downward
if it curves
down.
Warm-Up
Curve your pipe cleaner to fit the following
descriptions:
π is increasing
and concave
upward
π is decreasing
and concave
upward
π is increasing
and concave
downward
π is decreasing
and concave
downward
Warm-Up
In each case, where does the curve appear
in relation to its tangent lines?
π is increasing
and concave
upward
π is decreasing
and concave
upward
π is increasing
and concave
downward
π is decreasing
and concave
downward
Warm-Up
Curve your pipe cleaner to fit the following
descriptions:
π is increasing
at an
increasing rate
π is decreasing
at a decreasing
rate
π is increasing at
a decreasing
rate
π is decreasing
at an
increasing rate
Concavity
is related
to the rate
of change
of the rate
of change.
Objective 1
You will be able
concavity and
of the graph
to determine the
points of inflection
of a function
Exercise 1
Notice that the graph
of π(π₯) = π₯ 2 β 4π₯ + 3
is concave upward.
What is true about
the derivative of π?
What is true about
the second derivative
of π?
Definition of Concavity
Let π be differentiable on an open interval πΌ.
The graph of π lies
above all of its
tangent lines on πΌ.
The graph of π is
concave upward on
πΌ if πβ² is increasing
on the interval.
Definition of Concavity
Let π be differentiable on an open interval πΌ.
The graph of π is
concave downward
on πΌ if πβ² is decreasing
on the interval.
The graph of π lies
below all of its
tangent lines on πΌ.
Test for Concavity
Let π be a function whose second derivative
exists on an open interval πΌ.
If πβ²β²(π₯) > 0
for all π₯ in πΌ,
then the
graph of π is
concave
upward in πΌ.
If πβ²β²(π₯) < 0
for all π₯ in πΌ,
then the
graph of π is
concave
downward in
πΌ.
If πβ²β²(π₯) = 0
for all π₯ in πΌ,
then π is
linear and
has no
concavity.
Test for Concavity
Let π be a function whose second derivative
exists on an open interval πΌ.
If πβ²β²(π₯) < 0
If πβ²β²(π₯) > 0
If πβ²β²(π₯) = 0
for all π₯ in πΌ,
for all π₯is
in πΌ,
for all π₯πinis
πΌ,
This
similar tothen
testing
for where
the
then the
then π is
graph
of
π
is
increasing
or decreasing, except
graph
of π is
linear and
concavend
concave
has no
youβre using
the 2 in derivative.
downward
upward in πΌ.
concavity.
πΌ.
Exercise 2
Determine the open intervals which the
6
graph of π π₯ = 2 is concave upward or
π₯ +3
downward.
Exercise 3
Determine the open intervals on which the
π₯ 2 +1
π₯ 2 β4
graph of π π₯ =
is concave upward or
concave downward.
Points of Inflection
Let π be a function that is continuous
on an open interval and let π be a
point in the interval.
If the graph of π has a tangent line
at this point π, π π , then this
point is a point of inflection if the
concavity of π changes from
upward to downward (or downward
to upward) at the point.
Points of Inflection
Let π be a function that is continuous
on an open interval and let π be a
point in the interval.
If π, π π is a point of inflection
of the graph of π, then either
πβ²β²(π) = 0 or πβ²β² does not exist at
π₯ = π.
Exercise 4
Determine the points of
inflection and discuss the
concavity of the graph of
π(π₯) = π₯ 4 .
Exercise 5
Determine the points of inflection
and discuss the concavity of the
graph of π(π₯) = π₯ 4 β 4π₯ 3 .
You will be able to
apply the Second
Derivative Test to
find relative
extrema
Objective 2
Exercise 6a
Note that
1 3
π π₯ = π₯ β 2π₯ 2 + 3π₯ + 1
3
has a relative maximum
at π₯ = 1.
What is the value of the
second derivative at π₯ = 1?
Exercise 6b
Note that
1 3
π π₯ = π₯ β 2π₯ 2 + 3π₯ + 1
3
has a relative minimum at
π₯ = 3.
What is the value of the
second derivative at π₯ = 3?
Second Derivative Test
Let π be a function such that πβ²(π) = 0 and
the second derivative of π exists on an open
interval containing π.
If πβ²β²(π) > 0, then
π has a relative
minimum at
π, π π .
Second Derivative Test
Let π be a function such that πβ²(π) = 0 and
the second derivative of π exists on an open
interval containing π.
If π β²β² π < 0, then
π has a relative
maximum at
π, π π .
Exercise 7
Find the relative extrema for
π(π₯) = β3π₯ 5 + 5π₯ 3 .
Possible Max
Possible Min
π π₯
Increasing
Critical Points
Decreasing
Concave
Upward
πβ² π₯
+
0 or undefined
β
Increasing
πβ²β² π₯
+
β
+
Possible
Points of
Inflection
Concave
Downward
Decreasing
0 or
undefined
β
3-4: Concavity and the Second Derivative Test
Objectives:
Assignment:
1. To determine the
concavity and points
of inflection of the
graph of a function
β’ P. 195-197: 1-25 odd,
49-52, 76, 79-82
2. To apply the Second
Derivative Test to
find relative extrema
β’ P. 195: 17-39 odd