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SECOND DERIVATIVE TEST
Section 3.4
Calculus AP/Dual, Revised Β©2017
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 1
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FILL IN THE BLANK REVIEW
A. First Derivative Test and Concavity
1. πβ²positive then f is: _____________________________
2. πβ² negative then f is: _____________________________
3. πβ²β² positive then π is Concave _______
4. πβ²β² negative then π is Concave _______
5. πβ² π = π or πβ² π = π«π΅π¬ to which π is π of a critical point, then the point is called __________________
6. πβ²β²(π) = π or πβ²β²(π) = π«π΅π¬ and πβ²β²(π) changes signs, then the point is called _______________________
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 2
INCREASINGDECREASING
UP
DOWN
CRITICAL POINT
POINT OF INFLECTION
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Graph π π = ππ β π, πβ²(π), and πβ²β²(π)
REVIEW EXAMPLE
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 3
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Compare π π = ππ β π and πβ²β²(π)
REVIEW EXAMPLE
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 4
Opens
UP
Opens
DOWN
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Solve for πβ² π = π of π π = ππ β π and then take πβ²β²(π)
REVIEW EXAMPLE
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 5
( ) 3f x x x= β
( ) 2' 3 1f x x= β
23 1 0x β =
23 1x =
2 1
3x =
1
3
. .
x
C N
=
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Solve for πβ² π = π of π π = ππ β π and then take πβ²β²(π)
REVIEW EXAMPLE
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 6
1
3
. .
x
C N
=
( ) 2' 3 1f x x= β
( )'' 6f x x=
1 6"
3 3f
= +
1 6"
3 3f β = β
Relative
MAX
Relative
MIN
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Fβ VS Fββ
π β Test π ββ Test for Concavity
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 7
CRITICAL POINTS
ABSOLUTE/GLOBAL EXTREMA
RELATIVE/LOCAL MAX OR MIN
CONCAVITY
POINT OF INFLECTION
PROCESS F ββ = 0 OR DNE
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A. Let π be a function such that πβ² π = π and πβ²β² exists on an open interval containing π:
1. If πβ²β² π > π, then π(π) has a relative minimum at π, π π and π is concave
up
2. If π β²β²(π) < π, then π(π) has a relative maximum at π, π π and π is concave
down
3. If πβ² π = π and πβ²β² π = π, then π(π) is inconclusive. Use the first derivative test.
SECOND DERIVATIVE TEST
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 8
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STEPS
A. Identify all critical points using the first derivative.
B. Determine the second derivative
C. Plug in the critical points into the second derivative
1. If the critical point is positive, there is a relative minimum
2. If the critical point is negative, there is a relative maximum
3. If the critical point is zero, test is inconclusive and test must revert back to the first derivative test
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 9
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Using the Second Derivative Test to find the relative extrema of
π π = π +π
π.
EXAMPLE 1
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 10
( ) 12f x x xβ= +
( ) 2' 1 2f x xβ= β
2
2:1 0
take common denominator
CPx
β =
: 0, 2CP x =
( ) 2
2' 1
TAKE CRITICAL POINT
f xx
= β
2 22 0, 0x xβ = =
0, 2x =
2
2 2
20
x
x xβ =
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Using the Second Derivative Test to find the relative extrema of
π π = π +π
πand justify.
EXAMPLE 1
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 11
( )2
2' 1f x
x= β
( ) 3
3
4'' 4f x x
x
β= =
( )( )
( )
'' 2
'' 0
'' 2
f
f
f
β =
=
=
2β 2
.Inc
β
+
( )( )
( )( )
( )( )
3
3
3
4 4'' 2 ( )
2 22
4'' 0
0
4 4'' 2 ( )
2 22
f NEG
f Und
f POS
β = = = βββ
= =
= = = +Rel.
Min.
Rel.
Max.
0Inc.
.
Concave Down
Inc
Concave Up
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Using the Second Derivative Test to find the relative extrema of
π π = π +π
πand justify.
EXAMPLE 1
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 12
02β 2Rel.
Min.
Rel.
Max.
( )
( )
has a Relative Minimum at 2 when '' 0
has a Relative Maximum at 2 when '' 0
When = 0, the test is inconclusive and must use the
first derivative test to determine relative extrema.
f x x f
f x x f
x
=
= β
Inc.
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Using the Second Derivative Test to find the relative extrema of
π π = π +π
πand justify.
EXAMPLE 1
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 13
β β β β
β
β
β
β
x
y
( )
( )
has a Relative Minimum at 2 when '' 0
has a Relative Maximum at 2 when '' 0
When = 0, the test is inclusive and must use the
first derivative test to determine relative extrema.
f x x f
f x x f
x
=
= β
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Using the Second Derivative Test to find the relative extrema of
π π = βπππ + πππ and justify.
EXAMPLE 2
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 14
( ) 5 33 5f x x x= β +
( ) 4 2' 15 15f x x x= β +
( )2 2: 15 1 0CP x xβ β =
: 0, 1CP x x= =
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Using the Second Derivative Test to find the relative extrema of
π π = βπππ + πππ and justify.
EXAMPLE 2
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 15
( ) 4 2' 15 15f x x x= β +
( ) 3'' 60 30f x x x= β +
( )
( )
( )
'' 0
'' 1
'' 1
f
f
f
=
=
β =
0
30
30
β
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Using the Second Derivative Test to find the relative extrema of
π π = βπππ + πππ and justify.
EXAMPLE 2
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 16
01β 1Rel.
Min.
Rel.
Max.
( )If '' 0 =0, then we revert back
to the First Derivative Test to
decide the Relative Maxima.
f
( ) ( ) ( )5 3
1 3 1 5 1f β = β β + β
( )1 2f β = ( )1 2f = β
( ) ( ) ( )5 3
1 3 1 5 1f = β +
( ) ( )
( ) ( )
( )
has a Relative Minimum at 1, 2 when '' 0
has a Relative Maximum at 1,2 when '' 0
At " 0 0 it is inconclusive and must use the
first derivative test to determine relative extrema.
f x f
f x f
f
β β
=
( )
( )
( )
'' 0
'' 1
'' 1
f
f
f
=
=
β =
0
30
30
β
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Using the Second Derivative Test to find the relative extrema of
π π = βπππ + πππ and justify.
EXAMPLE 2
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 17
( ) ( )
( ) ( )
( )
has a Relative Minimum at 1, 2 when '' 0
has a Relative Maximum at 1,2 when '' 0
At " 0 0 it is inconclusive and must use the
first derivative test to determine relative extrema.
f x f
f x f
f
β β
=
β β β β
β
β
β
β
x
y
![Page 18: SECOND DERIVATIVE TEST - DANGMATH.com...Using the Second Derivative Test to find the relative extrema of π=β π + π and justify. EXAMPLE 2 7/30/2018 1:18 AM 3.4A: Second](https://reader033.vdocuments.site/reader033/viewer/2022051908/5ffb7238ae43110f320942f7/html5/thumbnails/18.jpg)
Using the Second Derivative Test to find the relative extrema of
π π =π
πππ β ππ β ππ and justify.
YOUR TURN
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 18
( ) ( )
( )
has a Relative Minimum at 3, 9 when '' 0
5 has a Relative Maximum at 1, when '' 0
3
f x f
f x f
β
β
ββ β β β β β β β β β
β
β
β
β
β
β
β
β
β
β
x
y
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Suppose that the function π has a continuous second derivative for all πand that π βπ = π, πβ² βπ = βπ, πβ²β² βπ = π. Let π be a function
whose derivative is given by πβ² π = ππ β πππ ΰ΅«
ΰ΅―
ππ π +
π πβ² π for all π. Write an equation of the tangent line to the graph of
π at the point of where π = βπ. Justify response.
EXAMPLE 3
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 19
( )
( )
1 2,
' 1 3
f
f
β =
β = β
( )1,2β
( )2 3 1y xβ = β +
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The figure below shows the graph of the derivative of π, πβ² on the closed interval βπ. π, π. π . The graph of πβ² has a horizontal tangent line at π = π and is linear on the interval βπ. π, π. ππ .
(a) Find the π-coordinates of the relative maxima of π. Justify your answer.
(b) Find the π-coordinates of the points of inflection of π. Justify your answer.
(c) On what intervals is π decreasing? Justify your answer.
(d) Is the function π twice-differentiable? Justify your answer.
EXAMPLE 4
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 20
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The figure below shows the graph of the derivative of π, πβ² on the closed interval βπ. π, π. π . The graph of πβ² has a horizontal tangent line at π = π and is linear on the interval βπ. π, π. ππ .
(a) Find the π-coordinates of the relative maxima of π. Justify your answer.
EXAMPLE 4A
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 21
( ) has a relative maximum at 1, 3
where ' changes sign from positive to negative
f x x x
f
= β =
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The figure below shows the graph of the derivative of π, πβ² on the closed interval βπ. π, π. π . The graph of πβ² has a horizontal tangent line at π = π and is linear on the interval βπ. π, π. ππ .
(b) Find the π-coordinates of the points of inflection of π. Justify your answer.
EXAMPLE 4B
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 22
( )
( )
has a POI at 0.75, 3.5 where ' changes from decreasing to increasing.
has a POI at 2,7 where ' changes from increasing to decreasing.
f f
f f
β
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The figure below shows the graph of the derivative of π, πβ² on the closed interval βπ. π, π. π . The graph of πβ² has a horizontal tangent line at π = π and is linear on the interval βπ. π, π. ππ .
(c) On what intervals is π decreasing? Justify your answer.
EXAMPLE 4C
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 23
( ) ( ) is decreasing where ' 0 at I 1,1 3,3.5 .f f I β
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The figure below shows the graph of the derivative of π, πβ² on the closed interval βπ. π, π. π . The graph of πβ² has a horizontal tangent line at π = π and is linear on the interval βπ. π, π. ππ .
(d) Is the function π twice-differentiable (Continuous and differentiable)? Justify your answer.
EXAMPLE 4D
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 24
is continuous and not differentiable because 0.75 has a sharp turn and therefore, not differentiable. f x =
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TO RECAP
A. The First Derivative tells us:
1. Extrema
2. Relative Maximum
3. Relative Minimum
B. The Second Derivative tells us:
1. Concavity
2. Points of Inflection
C. The Second Derivative TEST tells us:
1. Extrema
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 25
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Let π be twice differentiable. πβ² π > π and πβ²β² π > π for all reals. What is a possible value for π π if π π = π, π π = π and π π = π ?
(A) ππ
(B) π
(C) π
(D) ππ
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)
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Let π be twice differentiable. πβ² π > π and πβ²β² π > π for all reals. What is a possible value for π π if π π = π, π π = π and π π = π ?
AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)
7/30/2018 1:18 AM Β§3.4A: Second Derivative Test 27
Vocabulary Connections and Process Answer and Justifications
1st Derivative
Test
2nd Derivative
Test
( )' 0 : increasingh x
( )'' 0 : concave uph x
( ) ( ) ( )0,0 , 1,2 , 2,7
25 8
7 8 15
close to 14
+ =
D( ) ( )As ' 0 and '' 0, the function is
increasing and concave up. The only possible
answer is 14.
h x h x
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ASSIGNMENT
Page 192
31-40 all & Justify all responses
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