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October 25, 2014

October 25, 2014

Posted by **Michael** on Thursday, November 11, 2010 at 9:56pm.

- Linearization -
**drwls**, Thursday, November 11, 2010 at 10:11pmWhat is a? It does not appear in your formula for f(x)

Do you mean

f(x) = (sqrt16) - x or sqrt(16-x)?

The first case is trivial: 4 -x.

For the second case, use the Taylor series formula.

At x=0, the derivative of f(x) is

f'(x) =

(-1/2)/(16-x)^1/2 = -1/8

In linear approximation,

f(x) = f(0) + f'(0)*x

= 4 - (x/8)

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