McLaurin Polynomial for $e^{3x^{2}}$

Question

The Problem

My calculus book featured a few examples where Taylor Polynomials could be constructed from known McLaurin Polynomials. I wanted to try doing it on my own, so I made up my own problem, which was to find the first order Taylor Polynomial of $f$ where $f(x) = e^{3x^{2}}$ around $x = 1$.

First I calculated the answer (which I've double-checked already, and which is correct), using the standard method:

$$P_1(x) = f(1) + f'(1)(x-1)$$ $$P_1(x) = e^{3\times 1^{2}} + e^{3\times 1^{2}}6\times 1(x-1)$$ $$P_1(x) = e^3(1 + 6(x-1))$$

Then I tried getting the same result, using the methods developed in the book. However, as it seems, this gives a different result.

The Method

$$e^{3x^{2}} = e^{(3x^{2}-3)+3} = e^{3x^{2} -3}e^{3}$$

Now let $t = 3x^2 - 3$, which gives:

$$e^{3x^{2}} = e^3 e^t$$

$x = 1 \implies t=0$, so I tried replacing $e^t$ with its expansion around $x = 0$:

$$e^t = 1 + t +O(t^2)$$

If that's true, then:

$$P_1(x)=e^3(1 + t) = e^3(1 + 3(x^2 - 1))$$

Obviously, this cannot be true, since it contains an $x^2$-term, so I do see that it doesn't make sense. However, I want to ask a few questions.

My Questions

I'll list them, for simplicity:

1. Which part of the solution leads to the incorrect result? I'm guessing that it has to do with the fact that $t = 0$ doesn't imply that $x = 1$, since $x = -1$ also gives $t = 0$. Although I'm not sure.
2. Can I solve the problem correctly, using known McLaurin Polynomials? If so, how do I do that?
3. As an expansion of question 1: What exactly does what I calculated mean? Or does it even mean anything? Now when I think about it, I guess that (using the same arguments as in my solution) the polynomial $e^3(1 + 3(x^2 - 1))$ could be a "first order Taylor Polynomial" around $x = -1$ as well, which is really strange. Can I somehow understand what the heck I've done? Is there any meaning behind my result, whatsoever? Even if it's not important to the question, I'm curious about this, having sat down with it for a while now. (I could care less about the question itself, since I made it up...)
4. Maybe a redundant question since it's related to 3, but why do I get an $x^2$-term?

Lastly, I'm very tired, so if this is the result of me being uncareful somewhere with the maths, I am sorry about that. I've checked it several times to make sure that's not the case, but you never know what you're missing late in the evening.

Answer 1

Let us put $t=x-1$

or

$x=t+1$,

thus $e^{3x^2}=e^{3t^2+6t+3}$

$=e^3e^{3t^2+6t}$

$=e^3.(1+3t^2+6t+O(t^2))$

$=e^3.(1+6t+O(t^2))$

$=e^3(1+6(x-1))+O((x-1)^2).$

so no error.