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 * (25.2-25) 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) + (x-25)[f'(x) at x=25)

f'(x) = (1/2)*x^(-1/2)

f'(25) = (1/2)/5 = 1/10

Therefore

f(x) = 5 + (x-25)/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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