Taylor series expansion about $x = 3$

Find the Taylor series expansion of $f(x)=\sqrt{6x-x^2} $ about $x = 3$.

Looking through my notes, I think this could be solved with the binomial theorem. I am no longer sure binomial theorem is usable. I think I might have to generalize the form of the nth derivative, but I cannot find a pattern.

1 Answer

Write$$f(x) = \sqrt{6x-x^2} = \bigl[9 - (x-3)^2\bigr]^{1/2}$$and now you can apply the generalized binomial theorem:\begin{align*} \bigl[9 + (-(x-3)^2)\bigr]^{1/2} &= \sum_{k=0}^\infty \binom{1/2}{k} 9^{1/2-k} (- (x-3)^2)^k \\ &= \sum_{k=0}^\infty \binom{1/2}{k} 9^{1/2-k} (-1)^k (x-3)^{2k}. \end{align*}