1. **Problem:** Determine which of the functions $F_1(x) = \cos x$, $F_2(x) = -\cos x$, $F_3(x) = \sin x$, $F_4(x) = -\sin x$ is an antiderivative of $f(x) = -\cos x$ on $x \in (-\infty, +\infty)$. 2. **Formula and rule:** The antiderivative $F(x)$ of a function $f(x)$ satisfies $F'(x) = f(x)$. We need to find $F_i(x)$ such that $\frac{d}{dx}F_i(x) = -\cos x$. 3. **Check each candidate:** - $\frac{d}{dx}F_1(x) = \frac{d}{dx} \cos x = -\sin x \neq -\cos x$ - $\frac{d}{dx}F_2(x) = \frac{d}{dx} (-\cos x) = \sin x \neq -\cos x$ - $\frac{d}{dx}F_3(x) = \frac{d}{dx} \sin x = \cos x \neq -\cos x$ - $\frac{d}{dx}F_4(x) = \frac{d}{dx} (-\sin x) = -\cos x = f(x)$ 4. **Conclusion:** The function $F_4(x) = -\sin x$ is an antiderivative of $f(x) = -\cos x$.