Wat is de absolute extrema van f (x) = (x-2) (x-5) ^ 3 + 12in [1,4]?

Antwoord:

Lokale Minima. is #-2187/128.#

Global Minima#=-2187/128~=-17.09.#

Global Maxima #=64.#

Uitleg:

Voor extrema, #f '(x) = 0. #

#f '(x) = (x-2) * 3 (x-5) ^ 2 + (x-5) ^ 3 * 1 = (x-5) ^ 2 {3x-6 + x-5 = (4x-11) (x-5) ^ 2. #

#f '(x) = 0 rArr x = 5! in 1,4, # dus geen behoefte aan verdere cosideration & # X = 11/4 #

#f '(x) = (4x-11) (x-5) ^ 2, rArr f' '(x) = (4x-11) * 2 (x-5) + (x-5) ^ 2 * 4 = 2 (x-5) {4x-11 + 2x-10} = 2 (x-5) (6x-21). #

Nu, #f '' (04/11) 2 = (11 / 4-5) (33 / 2-21) = 2 (-9/4) (- 9/2)> 0, # dat laten zien, #f (11/4) = (11 / 4-2) (11 / 4-5) ^ 3 = (3/2) (- 9/4) ^ 3 = -2187/128, # is Lokale Minima.

Om Global Values te vinden, hebben we dat nodig #f (1) = (1-2) (1-5) ^ 3 = 64 # & #f (4) = (4-2) (4-5) ^ 3 = -2. #

Vandaar, Global Minima # = Min # {lokale minima, #f (1), f (4)} = min {-2187 / 128,64, -2} = min. {-17.09, 64, -2} = - 2187/128 ~ = -17.09 #

Global Maxima # = Max # {local maxima (wat niet bestaat), #f (1), f (4)} = max {64, -2} = 64. #