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 $x=2$?


1 Answer 1


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 ($x=2$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.