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I have a very easy question from the 2004 BC2 (Form B) AP Calculus exam. The question is:

$f$ is a function with derivatives of all orders for all real numbers. The third-degree Taylor polynomial for $f$ about $x=2$ is given by $$T(x)=7-9(x-2)^2-3(x-2)^3$$ b) [...] Determine whether $f(2)$ is a relative maximum, minimum, or neither, and justify your answer.

To solve this, I used the second derivative test for relative extremum:

Since $f'(2)=T'(2)=2$ and $f''(0)=T''(2)=-18<0$, $f$ has a relative maximum at $x=2$.

I was thinking, though, what if I used the first derivative test for relative extremum instead?

   x         2
     <–––––––|–––––––>
T'(x)    +   0   –

Since the first 3 derivatives of $f(x)$ equal those of $T(x)$ around $x=2$, and $T'(x)$ changes from positive to negative at $x=2$, $f$ has a relative maximum there.

Now first of all, of course the second method is longer. However, I think it may also be incorrect, or at least require further justification. I do not think, though, that further justification is needed since $f$ does indeed equal $T$ for those first three derivatives at $x=2$ by definition.

Of course, that last statement of mine may very well be wrong. So, my question is, what is wrong with the second method for determining that $f$ has a relative maximum at $x=2$?

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So apparently the reason is fairly simple:

The given Taylor Polynomial is not a series, so its interval of convergence is just the center—2 in this case.

Therefore, without further proof/explanation, one could not use $T’$ to determine if $f’$ is positive or negative at any point other than the center ($x=2$).

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