Using method of differentials?
2015-12-16 1:30 pm
Find the approximate change in f(x) = √(3+x^2) as x changes from x=1 to x= 1.02
回答 (2)
2015-12-16 1:35 pm
Let us find f(1.02)
f(1) = sqrt( 3+1^2) = sqrt(4) = 2
1.02-1 = 0.02
dx = 0.02
f'(x) = 2x / (2sqrt(3+x^2) )
f'(x) = x / sqrt(3+x^2)
f'(1) = 1 / sqrt(4) = 1/2
f(1.02) = f(1) + 0.02 f'(1)
f(1.02) = 2 + (0.02) (1/2)
= 2.01
As x changes from 1 to 1.02, f(x) changes from 2 to 2.01
The change is 0.01
f(1) = sqrt( 3+1^2) = sqrt(4) = 2
1.02-1 = 0.02
dx = 0.02
f'(x) = 2x / (2sqrt(3+x^2) )
f'(x) = x / sqrt(3+x^2)
f'(1) = 1 / sqrt(4) = 1/2
f(1.02) = f(1) + 0.02 f'(1)
f(1.02) = 2 + (0.02) (1/2)
= 2.01
As x changes from 1 to 1.02, f(x) changes from 2 to 2.01
The change is 0.01
2015-12-16 1:44 pm
let u = (3+x^2)
du/dx = 2x
f = √(3+x^2) = u^(1/2)
df/dx = (1/2) u^(-1/2) du/dx
= (1/2) 2x u^(-1/2)
= x u^(-1/2)
df = x/√(3+x^2) dx
Δf ~ x/(√(3+x^2) Δx
x=1
Δx= .02
Δf ~ 1/2 (.02) = .01
du/dx = 2x
f = √(3+x^2) = u^(1/2)
df/dx = (1/2) u^(-1/2) du/dx
= (1/2) 2x u^(-1/2)
= x u^(-1/2)
df = x/√(3+x^2) dx
Δf ~ x/(√(3+x^2) Δx
x=1
Δx= .02
Δf ~ 1/2 (.02) = .01
收錄日期: 2021-04-21 15:49:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151216053039AAXwS2S