If you think of it as approximating f(x)=sqrtx at x=128, you don't.
The Taylor polynomial, of degree n+1, to approximate f(x) centered at a is:
f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ . . . +(f^(n+1)(a))/((n+1)!)(x-a)^(n+1)
The function in the question is f(x)=sqrt(x) whose derivatives are:
f'(x)=1/(2sqrt(x)), f''(x)=-1/(4sqrt(x)^3), f'''(x)=3/(8sqrt(x)^5),
Consequently, if f(x)=sqrtx, then f'(a) does not exist for a=0 (nor do any higher-order derivatives).
However, approximating f(b) by a Taylor Polynomial centered at some a that is far from b isn't going to yield a good approximation anyway.
We need a way to think of sqrt128 as a function of a number much closer to 0.
Some thought and some experimentation and some arithmetical exploration are called for.
Eventually one may note that 128=2^7. What we are looking for then, is an approximation of the function f(x)=x^(7/2). The first 3 derivatives of this function do exist at 0.
Note that if a=0 is not specified by the exercise, choose a=1.
You can probably finish from here.
