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)=12a^{6}/x^{7}-12a^{12}/x^{13}

f''(x)=156a^{12}/x^{14}-84a^{6}/x^{8}

And evaluated at x=a:

f(a)=-1

f'(a)=0

f''(a)=72/a^{2}

This gives the expansion:

f(x)=36(x-a)^{2}/a^{2}-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

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