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
f(x) = 5 + (x-25)/10
f(25.2) = 5+ .02 = 5.02
There is nothing wrong with your answer. The correct value is
Maybe they wanted you to do the last step and compute an actual numerical approximation, 5.02.
I typed in 5.01999999999 and it took it. Weird.
Calculus Part 2 - y = tan(sqrtx) Find dy/dx. So I found it and I got the answer ...
Calculus - lim(x->infinity) sqrtx sin(1/sqrtx) Someone please start me off. I...
Calculus - Find the local linearization of g(x)=sqrt 4x near x=2 use local ...
Calculus - Find the linearization L(x) of y=e^(7x)ln(x) at a=1. L(x)=
calculus - Find the linearization L(x) of the function at a. f(x) = x3/4, a = 81
Calculus - Find the partial derivative for x. f(x,y)=1/SQRTx^2+y^2
calculus - find f(x -1) for the following function f(x) = sqrtx^2 - 3/ 6x + 6
Differential calculus - Given that f(x)=x, g(x)=x-1: h(x)=sqrtx-1. find f0g0h.
Linearization 2 - ind the linearization L(x) of the function at a = -1. f(x) = x...
Calculus - I don't know how to do the integral of e^(lnx^2)dx and the integral ...