F (x) = 4x ^ 2-2x + x / (x-1 / 4)의 지역 극한은 무엇입니까?

대답:

(2/3) + 3 + 2 ^ (5/3)) / 4 # (f) = f (1 / 4 + 2 ^ (- 5/3)

설명:

관찰, #f (x) = 4x ^ 2-2x + x / (x-1 / 4); × RR- {1/4}. #

# = 4x 2-2x + 1 / 4-1 / 4 + {(x-1 / 4) +1/4} / (x-1 / 4); xne1 / 4 #

# 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. #

자, 로컬 Extrema, #f '(x) = 0, # 과, "f_ (min) 또는 f_ (max)"에 따라 #f "(x)> 또는 <0"인 경우 "resp."#

#f '(x) = 0 #

#rArr4 {2 (x-1 / 4)} + 0 + 1 / 4 {(- 1) / (x-1 / 4) ^ 2} = 0 …

(x-1 / 4) ^ 3 = 1 / 32 = 2 ^ -5. #rArr8 (x-1 / 4) = 1 / {4 (x-1 / 4) ^ 2}

# rArr x = 1 / 4 + 2 ^ (- 5/3) #

더욱이, # (ast) rArr f "(x) = 8-1 / 4 {-2 (x-1 / 4) ^ - 3}"그래서, "#

(1 / 4 + 2 ^ (- 5/3)) = 8+ (1/2) (2 ^ (- 5/3)) ^ - 3> 0 #

# "그러므로,"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.#

그러므로, = (2 ^ (2/3) + 3 + 2 ^ (5/3)) / 4 # (1 / 4 + 2 ^ (- 5/3)

수학을 즐기세요.