Aplikasi Turunan_2nd Derivative

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Recap

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Cara menentukan nilai ekstrim mutlak:

1. Pastikan bahwa fungsi itu kontinyu pada interval

2. Temukan nilai f pada critical points dan endpoints/ titik ujung

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NILAI MAKSIMUM DAN MINIMUM

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Second Derivative Test for Local Extrema

Suppose that 𝑥 = 𝑐 is a critical point of 𝑓(𝑥) such that 𝑓′ 𝑐 = 0 and

that 𝑓′′ 𝑥 is continuous in a region around 𝑥 = 𝑐. Then,

1. If 𝑓′′ 𝑐 < 0 then 𝑥 = 𝑐 is a relative maximum.

2. If 𝑓′′ 𝑐 > 0 then 𝑥 = 𝑐 is a relative minimum.

3. If 𝑓′′ 𝑐 = 0 then 𝑥 = 𝑐 can be a relative maximum, relative

minimum or neither.

Calculus I 6

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Example 9

Use the second derivative test to classify the critical points of the function,

ℎ 𝑥 = 3𝑥5 − 5𝑥3 + 3

Solution:

ℎ′ 𝑥 = 15𝑥4 − 15𝑥2 = 15𝑥2 𝑥2 − 1 = 15𝑥2 𝑥 − 1 𝑥 + 1 ℎ′′ 𝑥 = 60𝑥3 − 30𝑥 = 30𝑥 2𝑥2 − 1

Critical points: 𝑥 = −1, 𝑥 = 0, and 𝑥 = 1

ℎ′′ −1 = −30, ℎ′′ 0 = 0, ℎ′′ 1 = 30

Calculus I 7

The Shape of a Graph

local maximum

local minimum not local

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Points of Inflection

𝑦 = 3 + sin 𝑥 𝑏𝑒𝑟𝑢𝑏𝑎ℎ 𝑘𝑒𝑐𝑒𝑘𝑢𝑛𝑔𝑎𝑛𝑛𝑦𝑎 𝑝𝑎𝑑𝑎 𝑡𝑖𝑡𝑖𝑘 (𝜋, 3)

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Pada x=0

𝑃𝑎𝑑𝑎 𝑥 = 0 𝑓′′ 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑡𝑒𝑟𝑑𝑒𝑓𝑖𝑛𝑖𝑠𝑖

Kecekungan berubah

𝑓′′ 𝑥 < 0 𝑝𝑎𝑑𝑎 𝑥 < 0 𝑑𝑎𝑛 𝑓′′ 𝑥 > 0 𝑝𝑎𝑑𝑎 𝑥 > 0

𝑓′′ 𝑥 𝑏𝑒𝑟𝑢𝑏𝑎ℎ 𝑡𝑎𝑛𝑑𝑎 𝑝𝑎𝑑𝑎 𝑥 = 0,

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𝑤𝑎𝑙𝑎𝑢𝑝𝑢𝑛 𝑓′′ 𝑥 = 0 𝑝𝑎𝑑𝑎 𝑥 = 0, 𝑡𝑎𝑛𝑑𝑎 𝑓′′ 𝑥 𝑡𝑖𝑑𝑎𝑘 𝑏𝑒𝑟𝑢𝑏𝑎ℎ,

𝑓′′ 𝑥 > 0 𝑢𝑛𝑡𝑢𝑘 𝑥 < 0 𝑑𝑎𝑛 𝑥 > 0 𝑦′′ = 12𝑥2

Tidak ada inflection point

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16

Kecekungan Fungsi

Fungsi 𝑓(𝑥) dikatakan cekung ke atas pada interval 𝐼 bila naik pada

interval 𝐼, dan 𝑓(𝑥) dikatakan cekung kebawah pada interval 𝐼 bila turun

pada interval I.

Teorema 6 Uji turunan kedua untuk kecekungan

1. Jika , maka f cekung ke atas pada I. 2. Jika , maka f cekung ke bawah pada I.

) ( ' x f ) ( ' x f

I

x

f

"

(

)



0

,





I

x

f

"

(

)



0

,





Grafik fungsi cekung keatas Grafik fungsi cekung kebawah

x y

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17 2 2 ) 2 ( 4 ) ( '    x x x x f 2 4 2 ) ( 2     x x x x f

Tentukan selang kecekungan dari

Contoh :

Jawab : 4 2 2

)

2

(

)

4

)(

2

(

2

)

2

)(

4

2

(

)

(

''





x

f

4 2

)

2

(

))

4

(

2

)

2

)(

4

2

)((

2

(





x

3 2 2

)

2

(

8

2

8

2













x

3

)

2

(

8





x

Grafik f cekung keatas pada

(

2

,



)

_{dan cekung kebawah pada}

selang (,2)

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18

c f(c)

(c,f(c)) titik belok

c f(c)

(c,f(c)) titik belok

Karena disebelah kiri c cekung keatas dan disebelah kanan c cekung kebawah

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19

c f(c)

(c,f(c)) bukan titik belok

Karena disekitar c tidak

Terjadi perubahan kecekungan

c

Walaupun di sekitar c

Terjadi perubahan

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20

1

2

)

(

.

1

f

x



x

3



4

)

(

.

2

f

x



x

Tentukan titik belok (jika ada) dari

2

6

)

(

'

x

f



,

f

''

(

x

)



12

x

●

0

+++++++ ---

Di x = 0 terjadi perubahan kecekungan, dan f(0)= -1 maka (0,-1) merupakan titik belok

2

12

)

(

''

x

f



●

0

+++++++ +++++++

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21

2

4

2

)

(

.

3

2









x

f

3

)

2

(

8

)

(

''





x

f

● 2 +++++ ---

Walaupun di x = 2, terjadi perubahan kecekungan, tidak ada

titik belok karena fungsi f(x) tidak terdefinisi di x = 2

f”(x)

x Tida

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Example 8

For the following function identify the intervals where the function is increasing and decreasing and the intervals where the function is concave up and concave down. Use this information to sketch the graph.

ℎ 𝑥 = 3𝑥5 − 5𝑥3 + 3

Solution:

ℎ′ 𝑥 = 15𝑥4 − 15𝑥2 = 15𝑥2 𝑥 − 1 𝑥 + 1 ℎ′′ 𝑥 = 60𝑥3 − 30𝑥 = 30𝑥 2𝑥2 − 1

Calculus I 22

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Example 8

ℎ 𝑥 = 3𝑥5 − 5𝑥3 + 3

Solution:

Calculus I 23

decreasing decreasing increasing

increasing

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Example 8

ℎ 𝑥 = 3𝑥5 − 5𝑥3 + 3

Solution:

Second derivative test for concavity:

ℎ′′ 𝑥 = 30𝑥 2𝑥2 − 1 = 0

𝑥 = 0, 𝑥 = ± 1

2 = ±0.7071

Calculus I 24

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Example 8

ℎ 𝑥 = 3𝑥5 − 5𝑥3 + 3

Solution:

Calculus I 25

concave up concave up

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Example 8

ℎ 𝑥 = 3𝑥5 − 5𝑥3 + 3

Solution:

Calculus I 26

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Second Derivative Test for Local Extrema

Suppose that 𝑥 = 𝑐 is a critical point of 𝑓(𝑥) such that 𝑓′ 𝑐 = 0 and

that 𝑓′′ 𝑥 is continuous in a region around 𝑥 = 𝑐. Then,

1. If 𝑓′′ 𝑐 < 0 then 𝑥 = 𝑐 is a relative maximum.

2. If 𝑓′′ 𝑐 > 0 then 𝑥 = 𝑐 is a relative minimum.

3. If 𝑓′′ 𝑐 = 0 then 𝑥 = 𝑐 can be a relative maximum, relative

minimum or neither.

Calculus I 27

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Example 9

Use the second derivative test to classify the critical points of the function,

ℎ 𝑥 = 3𝑥5 − 5𝑥3 + 3

Solution:

ℎ′ 𝑥 = 15𝑥4 − 15𝑥2 = 15𝑥2 𝑥2 − 1 = 15𝑥2 𝑥 − 1 𝑥 + 1 ℎ′′ 𝑥 = 60𝑥3 − 30𝑥 = 30𝑥 2𝑥2 − 1

ℎ′′ −1 = −30, ℎ′′ 0 = 0, ℎ′′ 1 = 30

Calculus I 28

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Exercises

For each of the functions in this exercise, answer each of the following.

a) Identify the critical points of the function.

b) Determine the open intervals on which the function increases and decreases.

c) Classify the critical points as relative maximums, relative minimums or neither.

d) Determine the open intervals on which the function is concave up and concave down.

e) Determine the inflection points of the function.

f) Use the information from steps (a) – (e) to sketch the graph of the function.

Calculus I 29

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Exercises

Calculus I 30

1.

𝑓 𝑥 = 10 − 30𝑥

2

+ 2𝑥

3

2.

𝑔 𝑤 = 𝑤 𝑤 − 4

1₃

3.

ℎ 𝑡 = 3𝑡 − 5 sin 2𝑡 on −1,4

4.

𝑓 𝑥 = 3𝑥𝐞

1−14𝑥2