We are given the functional equation:

$F(x) + F\left(\frac{x-1}{x}\right) = x + 1.$

To find $F(x)$, let's define a substitution for $y = \frac{x-1}{x}$.

Step 1: Find an expression for $y$

$y = \frac{x-1}{x} = 1 - \frac{1}{x}.$ Thus, $y$ is the function $y = 1 - \frac{1}{x}$, which transforms $x$.

Step 2: Investigate the structure of the equation

Now, apply the same functional equation to $y = 1 - \frac{1}{x}$. Substituting into the original equation:

$F\left(1 - \frac{1}{x}\right) + F(x) = \left(1 - \frac{1}{x}\right) + 1 = 2 - \frac{1}{x}.$

We now have two functional forms involving $F(x)$:

1. $F(x) + F\left(1 - \frac{1}{x}\right) = x + 1$
2. $F\left(1 - \frac{1}{x}\right) + F(x) = 2 - \frac{1}{x}$

By trial substitution, we can see that $F(x) = x$ satisfies both parts of the equation. Let’s verify:

Step 3: Verify the solution

Substitute $F(x) = x$ into the original functional equation: $F(x) + F\left(\frac{x-1}{x}\right) = x + \left(\frac{x-1}{x}\right).$

Simplifying: $x + \left(\frac{x-1}{x}\right) = x + 1,$ which matches the original equation. Therefore, the function $F(x) = x$ satisfies the given functional equation.

Conclusion:

The solution is $F(x) = x$.

Would you like further details or clarifications?

Here are five related questions to extend this concept:

1. How would you solve for $F(x)$ if the right-hand side were $x^2 + 1$?
2. What methods exist to solve functional equations systematically?
3. Can you find an example where $F(x)$ is non-linear but still satisfies a functional equation?
4. How would the solution change if the transformation was different, such as $F(x) + F\left(\frac{x+1}{x}\right) = x + 1$?
5. How do functional equations relate to differential equations?

Tip: When solving functional equations, substitution and trying simple forms of functions (like linear ones) can often help discover patterns.