Asked by Z32
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.
This is my math: 5+ 1/10 * (25.2-25) but it's wrong.
Answers
Answered by
drwls
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.
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.
Answered by
Z32
I typed in 5.01999999999 and it took it. Weird.
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