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 $x=2$?