If f is symmetrical about x = 1, find the real values of x satisfying the equation `f(x)=f((x+1)/(x+2))`.

Answer

To solve the equation \( f(x) = f\left(\frac{x+1}{x+2}\right) \) given that the function \( f \) is symmetrical about \( x = 1 \), we can follow these steps:

### Step 1: Understand the symmetry condition
Since \( f \) is symmetrical about \( x = 1 \), we have:
\[ f(1 + x) = f(1 - x) \]
This means that if we replace \( x \) with \( 1 + x \) or \( 1 - x \), the function values will be equal.

### Step 2: Set up the equation
We need to find \( x \) such that:
\[ f(x) = f\left(\frac{x+1}{x+2}\right) \]

### Step 3: Use the symmetry property
Using the symmetry property, we can set:
\[ x = 1 + t \quad \text{and} \quad \frac{x+1}{x+2} = 1 - t \]
This gives us:
\[ f(1 + t) = f(1 - t) \]
Thus, we need to solve:
\[ 1 + t = \frac{(1 + t) + 1}{(1 + t) + 2} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ (1 + t)(3 + t) = (2 + t)(1 + t) \]
Expanding both sides:
\[ 3 + 3t + t + t^2 = 2 + 2t + t + t^2 \]
This simplifies to:
\[ 3 + 4t = 2 + 3t \]

### Step 5: Solve for \( t \)
Rearranging the equation:
\[ 4t - 3t = 2 - 3 \]
This simplifies to:
\[ t = -1 \]

### Step 6: Substitute back to find \( x \)
Since \( t = -1 \):
\[ x = 1 + t = 1 - 1 = 0 \]

### Final Answer
The real value of \( x \) that satisfies the equation is:
\[ \boxed{0} \]