Question
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
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.
Z32
I typed in 5.01999999999 and it took it. Weird.