Odpowiedź:
# f_ (min) = f (1/4 + 2 ^ (- 5/3)) = (2 ^ (2/3) + 3 + 2 ^ (5/3)) / 4. #
Wyjaśnienie:
Obseruj to, #f (x) = 4x ^ 2-2x + x / (x-1/4); xw RR- {1/4}. #
# = 4x ^ 2-2x + 1 / 4-1 / 4 + {(x-1/4) +1/4} / (x-1/4); xne1 / 4 #
# = (2x-1/2) ^ 2-1 / 4 + {(x-1/4) / (x-1/4) + (1/4) / (x-1/4)}; xne1 / 4 #
# = 4 (x-1/4) ^ 2-1 / 4 + {1+ (1/4) / (x-1/4)}; xne1 / 4 #
#:. f (x) = 4 (x-1/4) ^ 2 + 3/4 + (1/4) / (x-1/4); xne1 / 4. #
Teraz, na Lokalne ekstrema, #f '(x) = 0, # i, #f '' (x)> lub <0, ”zgodnie z„ f_ (min) lub f_ (max), „resp.” #
#f '(x) = 0 #
#rArr 4 {2 (x-1/4)} + 0 + 1/4 {(- 1) / (x-1/4) ^ 2} = 0 … (ast) #
#rArr 8 (x-1/4) = 1 / {4 (x-1/4) ^ 2} lub, (x-1/4) ^ 3 = 1/32 = 2 ^ -5.
# rArr x = 1/4 + 2 ^ (- 5/3) #
Dalej, # (ast) rArr f '' (x) = 8-1 / 4 {-2 (x-1/4) ^ - 3}, „więc,” #
#f '' (1/4 + 2 ^ (- 5/3)) = 8+ (1/2) (2 ^ (- 5/3)) ^ - 3> 0 #
# „Dlatego” f_ (min) = f (1/4 + 2 ^ (- 5/3)) #
#=4(2^(-5/3))^2+3/4+(1/4)/(2^(-5/3))=2^2*2^(-10/3)+3/4+2^(-2)*2^(5/3)#
#=1/2^(4/3)+3/2^2+1/2^(1/3)=(2^(2/3)+3+2^(5/3))/4.#
A zatem, #f_ (min) = f (1/4 + 2 ^ (- 5/3)) = (2 ^ (2/3) + 3 + 2 ^ (5/3)) / 4. #
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