The problem is to determine whether the function $F(x) = \sin(x) - x + x^2$ is even, odd, or neither.

1. Problem Description

The problem is to determine whether the function $F(x) = \sin(x) - x + x^2$ is even, odd, or neither.

2. Solution Steps

To determine whether a function is even, odd, or neither, we need to analyze $F(-x)$.

An even function satisfies $F(-x) = F(x)$ for all $x$.

An odd function satisfies $F(-x) = -F(x)$ for all $x$.

If neither of these conditions is met, then the function is neither even nor odd.

Given $F(x) = \sin(x) - x + x^2$, we need to find $F(-x)$.

$F(-x) = \sin(-x) - (-x) + (-x)^2$

We know that $\sin(-x) = -\sin(x)$, so

$F(-x) = -\sin(x) + x + x^2$

Now, let's check if $F(-x) = F(x)$:

$F(x) = \sin(x) - x + x^2$

$F(-x) = -\sin(x) + x + x^2$

Since $\sin(x) - x + x^2 \ne -\sin(x) + x + x^2$, $F(x) \ne F(-x)$, so $F(x)$ is not an even function.

Next, let's check if $F(-x) = -F(x)$:

$-F(x) = -(\sin(x) - x + x^2) = -\sin(x) + x - x^2$

$F(-x) = -\sin(x) + x + x^2$

Since $-\sin(x) + x - x^2 \ne -\sin(x) + x + x^2$, $F(-x) \ne -F(x)$, so $F(x)$ is not an odd function.

Therefore, the function is neither even nor odd.

3. Final Answer

The function is neither odd nor even.