Approximate Expression from Taylor Expansion: f(x)=(a/x)^(12)-2(a/x)^6

[GonePhishing]

Given:

f(x)=(a/x)¹²-2(a/x)⁶
​

Use a Taylor Expansion about x=a to obtain an approximate expression for f(h), where h=x-a, to a second order in h.

So I've found:

f'(x)=12a⁶/x⁷-12a¹²/x¹³
f''(x)=156a¹²/x¹⁴-84a⁶/x⁸
​

And evaluated at x=a:

f(a)=-1
f'(a)=0
f''(a)=72/a²
​

This gives the expansion:

f(x)=36(x-a)²/a²-1
​

I thougt at this point, I could then do f(h) or f(x-a) by substituting x=h or x=x-a, but the answer is expected to be only in terms of h, so I am not sure what to do next.
Any help would be much appreciated

 

[HallsofIvy]

If the answer is "expected to be only in terms of h" then use h= x-a which is what the problem told you: \(\displaystyle \frac{36}{a^2}h^2- 1\).

 

[GonePhishing]

Which is what I thought, but I can't have an a term apparently.