Aplikasi Turunan_2nd Derivative

  Lecture 14

Recap

  • Nilai maksimum dan minimum
  • • Fungsi monoton naik dan turun

  Cara menentukan nilai ekstrim mutlak:

  1. Pastikan bahwa fungsi itu kontinyu pada interval

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

  3. Ambil nilai tertinggi dan terendah dari semua nilai yang ditemukan ini

  Fungsi naik dan Fungsi Turun

  Aplikasi Turunan Ke-Dua The Shape of a Graph Second Derivative Test for Local Extrema

  ′

  Suppose that = is a critical point of ( ) such that = 0 and

  ′′

  that is continuous in a region around = . Then,

  ′′

  1. If < 0 then = is a relative maximum.

  ′′

  2. If > 0 then = is a relative minimum.

  ′′

  3. If = 0 then = can be a relative maximum, relative minimum or neither. The Shape of a Graph Example 9

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

  

5

  3

  ℎ = 3 − 5 + 3

  Solution:

  4

  2

  2

  2

  2

  ℎ = 15 − 15 = 15 − 1 = 15 − 1 + 1

  ′′

  3

  2

  ℎ = 60 − 30 = 30 2 − 1 Critical points:

  = −1, = 0, and = 1

  ′′ ′′ ′′

  ℎ −1 = −30, ℎ 0 = 0, ℎ 1 = 30

  local not local local maximum maximum/minimum minimum

  Kecekungan / Concavity

  Points of Inflection = 3 + sin ℎ ( , 3) = ,

  ′ = cos = −1

  ′′ = − sin = 0 ,

  ′′ = 0

  ′′ Pada x=0 = 0

  ′′ Kecekungan berubah

  ′′ < 0 < 0

  ′′ > 0 > 0

  ′′ ℎ = 0, sehingga ada point of inflection

  ′′

  = 0 = 0,

  ′′ ℎ ,

  ′′ > 0 < 0 > 0

  ′′ = 12

  , ′′

  = 0 ′′ Kecekungan Fungsi y y x x

  Grafik fungsi cekung keatas Grafik fungsi cekung kebawah

  Fungsi

  f ' ( x ) f ' ( x )

  ( ) dikatakan cekung ke atas pada interval bila naik pada interval

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

  Teorema 6 Uji turunan kedua untuk kecekungan 1. Jika , maka f cekung ke atas pada I.

  " ( ) f x  ,  x

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

  " ( ) f x  ,  x

  I

  2 x

  2 x

  4

  f ( x )  Contoh :

  Tentukan selang kecekungan dari x

  2 

  Jawab : 2

  4 xx

  ' ( ) f x2

  ( 2 ) x2 2

  (

  2 4 )( 2 ) 2 ( 2 )( 4 ) xx   xxx

  '' ( ) f x4

  ( 2 ) x

  

  2 ( 2 )((

  2 4 )(

2 )

2 ( 4 )) x x x x x

       

  

4

( 2 ) x

  

  Tidak

  • ada f”(x)
  • 2 2<

    • 8

  2

  8

  8

  2

  8 xx   xx

  

  • 3 3

      x

      2

      ( x 2 )  ( 2 ) x

      Grafik f cekung keatas pada

      ( 2 , )  dan cekung kebawah pada

      ( , 2 ) selang  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

      Karena disebelah kiri c cekung kebawah dan disebelah kanan c cekung keatas f(c) c c

      (c,f(c)) bukan titik belok Walaupun di sekitar c Karena disekitar c tidak Terjadi perubahan Terjadi perubahan kecekungan Kecekungan tapi tidak ada

      Titik belok karena f tidak terdefinisi di c

      1 ( 2 ) .

      1

      3   x x f

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

      4 ) ( .

      2 x x f

       Tentukan titik belok (jika ada) dari

      2 ( 6 ) ' x x fx x f

      ( 12 ) '' , 

      ●

      2 ( 12 ) ''

       x x f

    • Tidak ada titik belok, karena tidak terjadi perubahan kecekungan f”(x) x f”(x) x

      ●

      2

      2

      4 xx

      3 . f ( x ) 

      2 x

      8 '' ( ) f x3

      ( 2 ) x

      Tida f”(x)

    • k

      ● ada x

      titik belok karena fungsi f(x) tidak terdefinisi di x = 2 The Shape of a Graph 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.

      

    5

      3

      ℎ = 3 − 5 + 3

      Solution:

      4

      2

      2

      ℎ = 15 − 15 = 15 − 1 + 1

      ′′

      3

      2

      ℎ = 60 − 30 = 30 2 − 1 Critical points:

      = −1, = 0, and = 1 The Shape of a Graph 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.

      

    5

      3

      ℎ = 3 − 5 + 3

      Solution:

    increasing decreasing decreasing increasing local local not local maximum maximum/minimum minimum The Shape of a Graph 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.

      

    5

      3

      ℎ = 3 − 5 + 3

      Solution:

      Second derivative test for concavity:

      ′′

      2

      ℎ = 30 2 − 1 = 0

      1 = 0, = ± = ±0.7071

      2 The Shape of a Graph 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.

      

    5

      3

      ℎ = 3 − 5 + 3

      Solution: concave down concave down concave up concave up The Shape of a Graph 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.

      

    5

      3

      ℎ = 3 − 5 + 3

      Solution: The Shape of a Graph Second Derivative Test for Local Extrema

      ′

      Suppose that = is a critical point of ( ) such that = 0 and

      ′′

      that is continuous in a region around = . Then,

      ′′

      1. If &lt; 0 then = is a relative maximum.

      ′′

      2. If &gt; 0 then = is a relative minimum.

      ′′

      3. If = 0 then = can be a relative maximum, relative minimum or neither. The Shape of a Graph Example 9

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

      

    5

      3

      ℎ = 3 − 5 + 3

      Solution:

      4

      2

      2

      2

      2

      ℎ = 15 − 15 = 15 − 1 = 15 − 1 + 1

      ′′

      3

      2

      ℎ = 60 − 30 = 30 2 − 1 Critical points:

      = −1, = 0, and = 1

      ′′ ′′ ′′

      ℎ −1 = −30, ℎ 0 = 0, ℎ 1 = 30

      local not local local maximum maximum/minimum minimum

    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. The Shape of a Graph

      Exercises The Shape of a Graph 1.

      = 10 − 30

      2

      3 2.

    • 2

      = − 4

      1

      3 3.

      ℎ = 3 − 5 sin 2 on −1,4 4.

      = 3 1−14

      2 5.

      = ln

      2

    • 1