SECOND DERIVATIVE TEST - DANGMATH.com...Using the Second Derivative Test to find the relative extrema of 𝒙=− 𝒙 + 𝒙 and justify. EXAMPLE 2 7/30/2018 1:18 AM 3.4A: Second

SECOND DERIVATIVE TEST

Section 3.4

Calculus AP/Dual, Revised ©2017

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7/30/2018 1:18 AM §3.4A: Second Derivative Test 1

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FILL IN THE BLANK REVIEW

A. First Derivative Test and Concavity

1. 𝒇′positive then f is: _____________________________

2. 𝒇′ negative then f is: _____________________________

3. 𝒇′′ positive then 𝒇 is Concave _______

4. 𝒇′′ negative then 𝒇 is Concave _______

5. 𝒇′ 𝒄 = 𝟎 or 𝒇′ 𝒄 = 𝑫𝑵𝑬 to which 𝒄 is 𝒙 of a critical point, then the point is called __________________

6. 𝒇′′(𝒙) = 𝟎 or 𝒇′′(𝒙) = 𝑫𝑵𝑬 and 𝒇′′(𝒙) changes signs, then the point is called _______________________


INCREASINGDECREASING

UP

DOWN

CRITICAL POINT

POINT OF INFLECTION

Graph 𝒇 𝒙 = 𝒙𝟑 − 𝒙, 𝒇′(𝒙), and 𝒇′′(𝒙)

REVIEW EXAMPLE


Compare 𝒇 𝒙 = 𝒙𝟑 − 𝒙 and 𝒇′′(𝒙)

REVIEW EXAMPLE


Opens

UP

Opens

DOWN

Solve for 𝒇′ 𝒄 = 𝟎 of 𝒇 𝒙 = 𝒙𝟑 − 𝒙 and then take 𝒇′′(𝒄)

REVIEW EXAMPLE


( ) 3f x x x= −

( ) 2' 3 1f x x= −

23 1 0x − =

23 1x =

2 1

3x =

1

3

. .

x

C N

=

Solve for 𝒇′ 𝒄 = 𝟎 of 𝒇 𝒙 = 𝒙𝟑 − 𝒙 and then take 𝒇′′(𝒄)

REVIEW EXAMPLE


1

3

. .

x

C N

=

( ) 2' 3 1f x x= −

( )'' 6f x x=

1 6"

3 3f

= +

1 6"

3 3f − = −

Relative

MAX

Relative

MIN

F’ VS F’’

𝒇 ’ Test 𝒇 ’’ Test for Concavity


CRITICAL POINTS

ABSOLUTE/GLOBAL EXTREMA

RELATIVE/LOCAL MAX OR MIN

CONCAVITY

POINT OF INFLECTION

PROCESS F ’’ = 0 OR DNE

A. Let 𝒇 be a function such that 𝒇′ 𝒄 = 𝟎 and 𝒇′′ exists on an open interval containing 𝒄:

1. If 𝒇′′ 𝒄 > 𝟎, then 𝒇(𝒙) has a relative minimum at 𝒄, 𝒇 𝒄 and 𝒇 is concave

up

2. If 𝒇 ′′(𝒄) < 𝟎, then 𝒇(𝒙) has a relative maximum at 𝒄, 𝒇 𝒄 and 𝒇 is concave

down

3. If 𝒇′ 𝒙 = 𝟎 and 𝒇′′ 𝒙 = 𝟎, then 𝒇(𝒙) is inconclusive. Use the first derivative test.

SECOND DERIVATIVE TEST


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


Using the Second Derivative Test to find the relative extrema of

𝒇 𝒙 = 𝒙 +𝟐

𝒙.

EXAMPLE 1


( ) 12f x x x−= +

( ) 2' 1 2f x x−= −

2

2:1 0

take common denominator

CPx

− =

: 0, 2CP x =

( ) 2

2' 1

TAKE CRITICAL POINT

f xx

= −

2 22 0, 0x x− = =

0, 2x =

2

2 2

20

x

x x− =


𝒇 𝒙 = 𝒙 +𝟐

𝒙and justify.

EXAMPLE 1


( )2

2' 1f x

x= −

( ) 3

3

4'' 4f x x

x

−= =

( )( )

( )

'' 2

'' 0

'' 2

f

f

f

− =

=

=

2− 2

.Inc

−

+

( )( )

( )( )

( )( )

3

3

3

4 4'' 2 ( )

2 22

4'' 0

0

4 4'' 2 ( )

2 22

f NEG

f Und

f POS

− = = = −−−

= =

= = = +Rel.

Min.

Rel.

Max.

0Inc.

.

Concave Down

Inc

Concave Up


𝒇 𝒙 = 𝒙 +𝟐

𝒙and justify.

EXAMPLE 1


02− 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

=

= −

Inc.


𝒇 𝒙 = 𝒙 +𝟐

𝒙and justify.

EXAMPLE 1


− − − −

−

−

−

−

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


f x x f

f x x f

x

=

= −


𝒇 𝒙 = −𝟑𝒙𝟓 + 𝟓𝒙𝟑 and justify.

EXAMPLE 2


( ) 5 33 5f x x x= − +

( ) 4 2' 15 15f x x x= − +

( )2 2: 15 1 0CP x x− − =

: 0, 1CP x x= =



EXAMPLE 2


( ) 4 2' 15 15f x x x= − +

( ) 3'' 60 30f x x x= − +

( )

( )

( )

'' 0

'' 1

'' 1

f

f

f

=

=

− =

0

30

30

−



EXAMPLE 2


01− 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 − = − − + −

( )1 2f − = ( )1 2f = −

( ) ( ) ( )5 3

1 3 1 5 1f = − +

( ) ( )

( ) ( )

( )

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


f x f

f x f

f

− −

=

( )

( )

( )

'' 0

'' 1

'' 1

f

f

f

=

=

− =

0

30

30

−



EXAMPLE 2


( ) ( )

( ) ( )

( )


has a Relative Maximum at 1,2 when '' 0

At " 0 0 it is inconclusive and must use the


f x f

f x f

f

− −

=

− − − −

−

−

−

−

x

y


𝒇 𝒙 =𝟏

𝟑𝒙𝟑 − 𝒙𝟐 − 𝟑𝒙 and justify.

YOUR TURN


( ) ( )

( )


5 has a Relative Maximum at 1, when '' 0

3

f x f

f x f

−

−

−− − − − − − − − − −

−

−

−

−

−

−

−

−

−

−

x

y

Suppose that the function 𝒇 has a continuous second derivative for all 𝒙and that 𝒇 −𝟏 = 𝟐, 𝒇′ −𝟏 = −𝟑, 𝒇′′ −𝟏 = 𝟓. Let 𝒈 be a function

whose derivative is given by 𝒈′ 𝒙 = 𝒙𝟒 − 𝟔𝒙𝟑 ൫

൯

𝟑𝒇 𝒙 +

𝟐 𝒇′ 𝒙 for all 𝒙. Write an equation of the tangent line to the graph of

𝒇 at the point of where 𝒙 = −𝟏. Justify response.

EXAMPLE 3


( )

( )

1 2,

' 1 3

f

f

− =

− = −

( )1,2−

( )2 3 1y x− = − +

The figure below shows the graph of the derivative of 𝒇, 𝒇′ on the closed interval −𝟏. 𝟓, 𝟑. 𝟓 . The graph of 𝒇′ has a horizontal tangent line at 𝒙 = 𝟐 and is linear on the interval −𝟏. 𝟓, 𝟎. 𝟕𝟓 .

(a) Find the 𝒙-coordinates of the relative maxima of 𝒇. Justify your answer.

(b) Find the 𝒙-coordinates of the points of inflection of 𝒇. Justify your answer.

(c) On what intervals is 𝒇 decreasing? Justify your answer.

(d) Is the function 𝒇 twice-differentiable? Justify your answer.

EXAMPLE 4



(a) Find the 𝒙-coordinates of the relative maxima of 𝒇. Justify your answer.

EXAMPLE 4A


( ) has a relative maximum at 1, 3

where ' changes sign from positive to negative

f x x x

f

= − =


(b) Find the 𝒙-coordinates of the points of inflection of 𝒇. Justify your answer.

EXAMPLE 4B


( )

( )

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

−


(c) On what intervals is 𝒇 decreasing? Justify your answer.

EXAMPLE 4C


( ) ( ) is decreasing where ' 0 at I 1,1 3,3.5 .f f I −


(d) Is the function 𝒇 twice-differentiable (Continuous and differentiable)? Justify your answer.

EXAMPLE 4D


is continuous and not differentiable because 0.75 has a sharp turn and therefore, not differentiable. f x =

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


Let 𝒉 be twice differentiable. 𝒉′ 𝒙 > 𝟎 and 𝒉′′ 𝒙 > 𝟎 for all reals. What is a possible value for 𝒉 𝟑 if 𝒉 𝟎 = 𝟎, 𝒉 𝟏 = 𝟐 and 𝒉 𝟐 = 𝟕 ?

(A) 𝟏𝟐

(B) 𝟕

(C) 𝟒

(D) 𝟏𝟒

AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)


Let 𝒉 be twice differentiable. 𝒉′ 𝒙 > 𝟎 and 𝒉′′ 𝒙 > 𝟎 for all reals. What is a possible value for 𝒉 𝟑 if 𝒉 𝟎 = 𝟎, 𝒉 𝟏 = 𝟐 and 𝒉 𝟐 = 𝟕 ?

AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)


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

ASSIGNMENT

Page 192

31-40 all & Justify all responses


Documents

SECOND DERIVATIVE TEST - DANGMATH.com...Using the Second Derivative Test to find the relative extrema of 𝒙=− 𝒙 + 𝒙 and justify. EXAMPLE 2 7/30/2018 1:18 AM 3.4A: Second