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(a) Given $f(x) = \sqrt{x}$, find its Taylor polynomial of degree 2 centered at $x=4$ and use it to estimate $\sqrt{5}$.

(b) Use Taylor's theorem to give an upper error bound for the estimate in part (a).

So the expansion from (a) is $P_{2}(x)=2+\frac{1}{4}(x-4)-\frac{1}{64}(x-4)^{2}$

and

$P_{2}(5)=2+\frac{1}{4}(5-4)-\frac{1}{64}(5-4)^{2}=\frac{145}{64}$

I have tried using the error formula $E_{n}(x)\leq\frac{M}{(n+1)!}(x-4)^{n+1}$ where $M=max(f^{n+1}(c))$ on $[4,5]$

The problem is my supposed max error is less than the error of $\sqrt{5}-\frac{143}{64}$

How do I determine what to use for $c$ so that my error makes sense?

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    $\begingroup$ You have a sign error in $P_2(x)$. Fix that, and you should be good. $\endgroup$ – Paul Sinclair Nov 23 '15 at 0:03
  • $\begingroup$ Is said error the sign of $\frac{1}{64}$? $\endgroup$ – MC989 Nov 23 '15 at 0:14
  • $\begingroup$ That is correct. $\endgroup$ – Paul Sinclair Nov 23 '15 at 2:52
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Correcting the error made in (a) to make $\frac{1}{64}$ negative (I already edited the question as such), then simply use $c=4$ (since this is where the $M=max(f^{n+1}(c))$ on $[4,5]$ occurs) and solve. Then the two errors are compatible if you compare them.

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