1. **State the problem:** Determine whether the function $f(x) = e^{-x^2} \cos x$ is even, odd, or neither. 2. **Recall definitions:** - A function $f$ is **even** if $f(-x) = f(x)$ for all $x$. - A function $f$ is **odd** if $f(-x) = -f(x)$ for all $x$. 3. **Evaluate $f(-x)$:** $$f(-x) = e^{-(-x)^2} \cos(-x) = e^{-x^2} \cos(-x)$$ 4. **Use properties of cosine:** Since cosine is an even function, $\cos(-x) = \cos x$, so $$f(-x) = e^{-x^2} \cos x$$ 5. **Compare $f(-x)$ and $f(x)$:** We see that $$f(-x) = e^{-x^2} \cos x = f(x)$$ 6. **Conclusion:** Since $f(-x) = f(x)$, the function $f(x) = e^{-x^2} \cos x$ is an **even function**.