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.