Kaj so lokalni ekstremi f (x) = x ^ 2 (x + 2)?

Odgovor:

Minima #(0, 0)#

Maxima #(-4/3, 1 5/27)#

Pojasnilo:

Glede na -

# y = x ^ 2 (x + 2) #

# y = x ^ 3 + 2x ^ 2 #

# dy / dx = 3x ^ 2 + 4x #

# (d ^ 2y) / (dx ^ 2) = 6x + 4 #

# dy / dx = 0 => 3x ^ 2 + 4x = 0 #

#x (3x + 4) = 0 #

# x = 0 #

# 3x + 4 = 0 #

# x = -4 / 3 #

At # x = 0; (d ^ 2y) / (dx ^ 2) = 6 (0) + 4 = 4> 0 #

At # x = 0; dy / dx = 0; (d ^ 2y) / (dx ^ 2)> 0 #

Funkcija ima zato minimalne vrednosti # x = 0 #

At # x = 0; y = (0) ^ 2 (0 + 2) = 0 #

Minima #(0, 0)#

At # x = -4 / 3; (d ^ 2y) / (dx ^ 2) = 6 (-4/3) + 4 = -4 <0 #

At # x = -4; dy / dx = 0; (d ^ 2y) / (dx ^ 2) <0 #

Funkcija ima torej maksimum na # x = -4 / 3 #

At # x = -4 / 3; y = (- 4/3) ^ 2 (-4 / 3 + 2) = 1 5/27 #

Maxima #(-4/3, 1 5/27)#

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