Problem 3. (6 points) 1. Find the first four nonzero terms in the MacLaurin series for the function $f(x)=e^{-x^{2}}$ A. $-x^{2}-x^{3}+\frac{x^{4}}{2}-\frac{x^{6}}{3 !}$ B. $1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{3 !}$ C. $1+\frac{x^{2}}{2}+\frac{x^{4}}{4 !}+\frac{x^{6}}{6 !}$ D. none of these E. $1+x^{2}+\frac{x^{4}}{2}+\frac{x^{6}}{3 !}$

Step 1 ：The problem is asking for the first four nonzero terms in the MacLaurin series for the function \(f(x)=e^{-x^{2}}\).

Step 2 ：The MacLaurin series for a function f(x) is the representation of the function as an infinite sum of terms that are calculated from the values of the function's derivatives at zero.

Step 3 ：The MacLaurin series for the function \(e^x\) is \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...\).

Step 4 ：We can derive the MacLaurin series for the function \(f(x)=e^{-x^{2}}\) by substituting \(-x^2\) for \(x\) in the series for \(e^x\).

Step 5 ：The first four nonzero terms of this series are \(1 - x^{2} + \frac{x^{4}}{2}\). The term \(O(x^6)\) represents the remainder of the series and is not considered as one of the first four terms.

Step 6 ：Therefore, the correct answer is \(\boxed{B. 1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{3 !}}\)