Posted by Z32 on Sunday, September 27, 2009 at 11:37pm.
Find the linearization of f(x)= sqrtx at x=25 and use the linearization to approximate sqrt25.2
This is my math: 5+ 1/10 * (25.225) but it's wrong.

Calculus  drwls, Monday, September 28, 2009 at 6:17am
The McLaurin series expansion of that function about x=25 is
f(x) = f(25) + (x25)[f'(x) at x=25)
f'(x) = (1/2)*x^(1/2)
f'(25) = (1/2)/5 = 1/10
Therefore
f(x) = 5 + (x25)/10
f(25.2) = 5+ .02 = 5.02
There is nothing wrong with your answer. The correct value is
5.01996016...
Maybe they wanted you to do the last step and compute an actual numerical approximation, 5.02.

Calculus  Z32, Monday, September 28, 2009 at 6:13pm
I typed in 5.01999999999 and it took it. Weird.
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