1. **Problem Statement:** Determine if the function $f(x) = 3e^{1-x} - 2\sin(\pi x) + x$ is one-to-one. 2. **Recall the definition:** A function is one-to-one (injective) if for every $x_1 \neq x_2$, we have $f(x_1) \neq f(x_2)$. Equivalently, the function must be strictly increasing or strictly decreasing over its entire domain. 3. **Check the derivative:** To test monotonicity, find $f'(x)$: $$ f'(x) = \frac{d}{dx}\left(3e^{1-x} - 2\sin(\pi x) + x\right) = 3 \cdot (-1) e^{1-x} - 2 \cdot \pi \cos(\pi x) + 1 = -3e^{1-x} - 2\pi \cos(\pi x) + 1 $$ 4. **Analyze $f'(x)$:** - The term $-3e^{1-x}$ is always negative because $e^{1-x} > 0$. - The term $-2\pi \cos(\pi x)$ oscillates between $-2\pi$ and $2\pi$ because $\cos(\pi x)$ ranges from $-1$ to $1$. - The constant $+1$ shifts the derivative. 5. **Check if $f'(x)$ can change sign:** Since $-3e^{1-x}$ is negative and $-2\pi \cos(\pi x)$ oscillates, $f'(x)$ will oscillate and can be positive or negative depending on $x$. 6. **Conclusion:** Because $f'(x)$ is not always positive or always negative, $f(x)$ is not strictly monotonic. Therefore, $f(x)$ is **not one-to-one**. **Final answer:** The function $f(x) = 3e^{1-x} - 2\sin(\pi x) + x$ is not one-to-one.