matematiksel iktisat ders notları (optimizasyon)

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77

Örnek 15:

y = f ( x ) = x6 + 5

f ′ ( x ) = 6 x 5 = 0 → x0 = 0

5 ′ f ( x) = 6x

f ′ ( 0) = 0

f ′′ ( x ) = 30 x 4

f ′′ ( 0 ) = 0

3 ′′′ f ( x ) = 120 x

f ′′′ ( 0 ) = 0

f

( 4)

( x ) = 360 x

f

( 5)

( x ) = 720 x

f ( 6) ( x ) = 720 x0 = 0 , y = 5

2

f

( 4)

( 0) = 0

f

( 5)

( 0) = 0

f ( 6) ( 0 ) = 720 > 0

noktasında minimum var.


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