If `f(x)=(x+1)^(2)-1`, then `f(x)=x` give `x` as

Options

A. `{-1,1}`
B. `{0,-1}`
C. `{0,1}`
D. An equivalence relation

Answer

To solve the equation \( f(x) = x \) where \( f(x) = (x+1)^2 - 1 \), we will follow these steps:

### Step 1: Set up the equation
We start with the function:
\[ f(x) = (x + 1)^2 - 1 \]
We need to find \( x \) such that:
\[ f(x) = x \]
This gives us the equation:
\[ (x + 1)^2 - 1 = x \]

### Step 2: Simplify the equation
Next, we simplify the equation:
\[ (x + 1)^2 - 1 - x = 0 \]
Expanding \( (x + 1)^2 \):
\[ x^2 + 2x + 1 - 1 - x = 0 \]
This simplifies to:
\[ x^2 + 2x - x = 0 \]
or
\[ x^2 + x = 0 \]

### Step 3: Factor the equation
Now we can factor the equation:
\[ x(x + 1) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor equal to zero gives us:
1. \( x = 0 \)
2. \( x + 1 = 0 \) which simplifies to \( x = -1 \)

### Step 5: Conclusion
Thus, the solutions for \( x \) are:
\[ x = 0 \quad \text{and} \quad x = -1 \]

### Final Answer
The values of \( x \) that satisfy \( f(x) = x \) are \( x = 0 \) and \( x = -1 \).

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