Key Concepts and Formulas

• Functional Equations: Equations involving unknown functions. Solving them often involves strategic substitution of variables.
• Cyclic Transformations: Recognizing patterns where repeated application of a transformation returns the original variable. This leads to systems of equations.
• System of Linear Equations: Solving a set of equations simultaneously to find the values of unknown variables.

Step-by-Step Solution

Problem Statement: We are given the functional equation: $f(x)+f\left(\frac{1}{1-x}\right)=1+x \quad (*)$ defined for $x \in \mathbb{R} - \{0,1\}$. We need to find the value of $f(2)$.

Step 1: Substitution to Introduce $f(2)$ To find $f(2)$, we substitute $x=2$ into the given functional equation $(*)$. The value $x=2$ is valid as it is not $0$ or $1$. $f(2)+f\left(\frac{1}{1-2}\right)=1+2$ $f(2)+f\left(\frac{1}{-1}\right)=3$ $f(2)+f(-1)=3 \quad \text{(Equation 1)}$ Reasoning: This substitution directly incorporates the term $f(2)$ we want to find. It also introduces $f(-1)$, suggesting that a substitution with $x=-1$ might be beneficial in the next step.

Step 2: Substitution to Relate $f(-1)$ From Equation 1, we have a term $f(-1)$. We substitute $x=-1$ into the original functional equation $(*)$. The value $x=-1$ is valid as it is not $0$ or $1$. $f(-1)+f\left(\frac{1}{1-(-1)}\right)=1+(-1)$ $f(-1)+f\left(\frac{1}{2}\right)=0 \quad \text{(Equation 2)}$ Reasoning: This step establishes a relationship between $f(-1)$ and $f(1/2)$, allowing us to connect the terms involved in the functional equation through a sequence of substitutions.

Step 3: Substitution to Close the Cycle From Equation 2, we have a term $f(1/2)$. We substitute $x=1/2$ into the original functional equation $(*)$. The value $x=1/2$ is valid as it is not $0$ or $1$. $f\left(\frac{1}{2}\right)+f\left(\frac{1}{1-\frac{1}{2}}\right)=1+\frac{1}{2}$ $f\left(\frac{1}{2}\right)+f\left(\frac{1}{\frac{1}{2}}\right)=\frac{3}{2}$ $f\left(\frac{1}{2}\right)+f(2)=\frac{3}{2} \quad \text{(Equation 3)}$ Reasoning: This crucial substitution reintroduces $f(2)$, which was present in Equation 1. This indicates that the transformation $x \mapsto \frac{1}{1-x}$ applied sequentially ($2 \mapsto -1 \mapsto 1/2 \mapsto 2$) creates a cycle of three distinct values, allowing us to form a solvable system of linear equations.

Step 4: Solving the System of Linear Equations We now have a system of three linear equations involving $f(2)$, $f(-1)$, and $f(1/2)$:

1. $f(2)+f(-1)=3$
2. $f(-1)+f\left(\frac{1}{2}\right)=0$
3. $f\left(\frac{1}{2}\right)+f(2)=\frac{3}{2}$

We can solve this system to find $f(2)$. From Equation 2, we can express $f(-1)$ in terms of $f(1/2)$: $f(-1) = -f\left(\frac{1}{2}\right)$

Substitute this into Equation 1: $f(2) + \left(-f\left(\frac{1}{2}\right)\right) = 3$ $f(2) - f\left(\frac{1}{2}\right) = 3 \quad \text{(Equation 4)}$

Now we have a simpler system with two equations and two unknowns ($f(2)$ and $f(1/2)$): 3. $f\left(\frac{1}{2}\right)+f(2)=\frac{3}{2}$ 4. $-f\left(\frac{1}{2}\right)+f(2)=3$

Add Equation 3 and Equation 4: $\left(f\left(\frac{1}{2}\right)+f(2)\right) + \left(-f\left(\frac{1}{2}\right)+f(2)\right) = \frac{3}{2} + 3$ $2f(2) = \frac{3}{2} + \frac{6}{2}$ $2f(2) = \frac{9}{2}$

Divide by 2 to find $f(2)$: $f(2) = \frac{9}{2} \times \frac{1}{2}$ $f(2) = \frac{9}{4}$ Reasoning: By systematically eliminating variables through substitution, we isolate the value of $f(2)$. The cyclical nature of the transformation ensured that we could form a closed system of equations.

Common Mistakes & Tips

• Domain Restrictions: Always verify that the substituted values of $x$ are within the function's domain ($\mathbb{R} - \{0,1\}$). For this problem, $2, -1, 1/2$ are all valid.
• Cyclic Property Identification: The transformation $g(x) = \frac{1}{1-x}$ has a period of 3, meaning $g(g(g(x))) = x$. Recognizing this cyclic property is key to setting up the system of equations efficiently.
• Systematic Solving: When solving the system of equations, ensure all arithmetic is accurate, especially with fractions and signs. Using substitution to simplify the system before adding/subtracting can prevent errors.

Summary

The problem involves a functional equation where the arguments are related by the transformation $x \mapsto \frac{1}{1-x}$. By strategically substituting values that form a cycle under this transformation ($2 \to -1 \to 1/2 \to 2$), we generated a system of three linear equations. Solving this system by substitution and elimination allowed us to determine the value of $f(2)$.

The final answer is $\boxed{\frac{9}{4}}$.