Yahoo Answers: Answers and Comments for Use a linear approximation (or differentials) to estimate the given number. (Use the linearization of 1/x. Do? [Mathematics]
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From Anonymous
enUS
Sat, 12 Mar 2011 03:10:02 +0000
3
Yahoo Answers: Answers and Comments for Use a linear approximation (or differentials) to estimate the given number. (Use the linearization of 1/x. Do? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20110312031002AAuv3wK
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From mathsmanretired: f(x) = 1/x > f '(x) = 1/x^2
1/99 ...
https://answers.yahoo.com/question/index?qid=20110312031002AAuv3wK
https://answers.yahoo.com/question/index?qid=20110312031002AAuv3wK
Sat, 12 Mar 2011 11:33:30 +0000
f(x) = 1/x > f '(x) = 1/x^2
1/99 is approximately 1/100 so close to this the gradient of the tangent is
1/(100^2) = 0.0001
This means that for every unit to the right the function decreases by 0.0001 approximately and so for every unit to the left it increases by 0.0001 approximately.
Therefore 1/99 =approx= 1/100 + 0.0001 = 0.0101
Edit. A calculator will of course confirm that 1/99 = 0.010101010101.......

From Anonymous: metode 1)
f(x)=1/x
f '(x)= 1/x²
f(x + ∆x)...
https://answers.yahoo.com/question/index?qid=20110312031002AAuv3wK
https://answers.yahoo.com/question/index?qid=20110312031002AAuv3wK
Sat, 12 Mar 2011 11:21:53 +0000
metode 1)
f(x)=1/x
f '(x)= 1/x²
f(x + ∆x) ≈ f(x) + ∆x f '(x)
we have:
x= 100
∆x= 1
x + ∆x = 100 + (1) = 99
f(100)= 1/100 = 0.01
f '(100) = 1/100² = 0.0001
hence
f(x + ∆x) ≈ f(x) + ∆x f '(x)
f(100 + (1)) ≈ f(100) + (1) f '(100)
f(99) ≈ 1/100 +(1)(1/100² )
f(99) ≈ 0.01 +(1)(0.0001)
1/99 ≈ 0.01 + 0.0001
1/99 ≈ 0.0101
metode 2)
f(x)=1/x
1/99 = 1/(100  1)
1/99 = (100  1)^(1) : Binomial expansion
1/99 = 100^(1) + (1)(1)100^(2) + (1)(2)/2 (1)^2 100^(3) + . . .
1/99 = 1/100 + 1/100^(2) + 1/100^(3) + . . .
1/99 = 0.01 + 0.0001 + 0.000001 + . . .
1/99 = 0.0101010 . . .