If I approximate $\displaystyle e^{-x^2}$ with the first three terms of its Maclaurin series, how do I find the range of x for which the approximation is accurate to within .005?

First, I found the Maclaurin series that represents this function:

$\displaystyle e^{-x^2} = \sum_{n=0}^\infty\frac{(-x^2)^n}{n!} = \sum_{n=0}^\infty\frac{x^{2n}}{n!} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + ...$

Since we are approximating using three terms, I have to find the 4th derivative of the function, which is $\displaystyle e^{-x^2}(16x^4-48x^2+12)$.

Up to this point, I think that I am in good shape. Even this next step seems fairly straightforward:

$\displaystyle |error| = |R_3(x,z)| = |\frac{f^{(4)}(z)}{4!}x^4| \leq 0.005$

It's in executing this that get confused. I think that the next step is this:

$\displaystyle |\frac{e^{-x^2}(16x^4-48x^2+12)}{24}x^4| \leq 0.005$

$\displaystyle |4x^4e^{-x^2}(4x^4-12x^2+3)| \leq 0.12$

It's at this point where I get stuck. Assuming that I have gotten this far without error (big assumption), the idea would be to solve the inequality for x, which would tell me the range to fall within this error threshold. But the algebra here seems pretty involved, which makes me think that I have made a mistake.

Anybody see anything or have any advice? Thanks.