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 \( f \) is symmetrical about \( x = 1 \), we can follow these steps:

### Step 1: Set up the equation
Since \( f \) is symmetrical about \( x = 1 \), we can use the property of symmetry. This means that:
\[ f(1 - x) = f(1 + x) \]
We need to find the values of \( x \) that satisfy:
\[ f(x) = f\left(\frac{x+1}{x+2}\right) \]

### Step 2: Equate the arguments
For the equality \( f(x) = f\left(\frac{x+1}{x+2}\right) \) to hold, the arguments must be equal due to the symmetry property:
\[ x = \frac{x+1}{x+2} \]

### Step 3: Cross-multiply
Cross-multiplying gives:
\[ x(x + 2) = x + 1 \]
This simplifies to:
\[ x^2 + 2x = x + 1 \]

### Step 4: Rearrange the equation
Rearranging the equation results in:
\[ x^2 + 2x - x - 1 = 0 \]
which simplifies to:
\[ x^2 + x - 1 = 0 \]

### Step 5: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = -1 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 6: Find the real values of \( x \)
Thus, the solutions from the quadratic equation are:
\[ x = \frac{-1 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{-1 - \sqrt{5}}{2} \]

### Step 7: Consider the symmetry condition
Since \( f \) is symmetrical about \( x = 1 \), we also need to check if there are other solutions. We can also check the condition:
\[ f(1 - x) = f(1 + x) \]
This implies that if \( x = 0 \), then \( f(1) = f(1) \) holds true. Thus, \( x = 0 \) is also a solution.

### Final Solutions
The real values of \( x \) satisfying the equation are:
\[ x = \frac{-1 + \sqrt{5}}{2}, \quad x = \frac{-1 - \sqrt{5}}{2}, \quad \text{and} \quad x = 0 \]