1. The problem is to analyze the function: $$F(x) = f(a+xh) + f(a-xh)$$ where $f$ is some function and $a, h$ are constants. 2. Step one is to understand that $F(x)$ is composed by evaluating $f$ at two points symmetric about $a$: at $a+xh$ and $a-xh$. 3. If $f$ is differentiable, we can expand $f(a+xh)$ and $f(a-xh)$ using Taylor series around $a$ to better understand the behavior of $F(x)$: $$f(a+xh) = f(a) + (xh)f'(a) + \frac{(xh)^2}{2}f''(a) + \cdots$$ $$f(a-xh) = f(a) - (xh)f'(a) + \frac{(xh)^2}{2}f''(a) - \cdots$$ 4. Adding these two expansions, the linear terms $(xh)f'(a)$ cancel out, so: $$F(x) = f(a+xh) + f(a-xh) = 2f(a) + (xh)^2 f''(a) + \text{higher even order terms}$$ 5. This shows $F(x)$ is an even function about $x = 0$ and its behavior depends mainly on $f$'s second derivative at $a$ at small $x$. 6. If $f$ is known, $F(x)$ can be computed or analyzed further for specific $x$ values. Final answer: $$F(x) = f(a+xh) + f(a-xh)$$ combines the values of $f$ at points symmetric around $a$, and its Taylor expansion shows the first order terms cancel, leaving even-powered terms important for the function's shape around $x=0$.