Paggamit ng kaugalian, hanapin ang humigit-kumulang na halaga ng (0.009) ^ (1/3)?

Paggamit ng kaugalian, hanapin ang humigit-kumulang na halaga ng (0.009) ^ (1/3)?
Anonim

Sagot:

#0.02083# (totoong halaga #0.0208008#)

Paliwanag:

Maaari itong malutas sa formula ng Taylor:

#f (a + x) = f (a) + xf '(a) + (x ^ 2/2) f' '(a) …. #

Kung #f (a) = a ^ (1/3) #

Magkakaroon kami ng:

#f '(a) = (1/3) a ^ (- 2/3) #

ngayon kung # a = 0.008 # pagkatapos

#f (a) = 0.2 # at

#f '(a) = (1/3) 0.008 ^ (- 2/3) = 25/3 #

Kaya kung # x = 0.001 # pagkatapos

#f (0.009) = f (0.008 + 0.001) ~~ f (0.008) + 0.001xxf '(0.008) = #

#=0.2+0.001*25/3=0.2083#