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 can follow these steps:

### Step 1: Set the equation
We start with the given function and set it equal to \( x \):
\[ (x + 1)^2 - 1 = x \]

### Step 2: Expand the left side
Next, we expand the left side using the formula for the square of a binomial:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
So, substituting this back into our equation gives:
\[ x^2 + 2x + 1 - 1 = x \]
This simplifies to:
\[ x^2 + 2x = x \]

### Step 3: Rearrange the equation
Now, we rearrange the equation to set it to zero:
\[ x^2 + 2x - x = 0 \]
This simplifies to:
\[ x^2 + x = 0 \]

### Step 4: Factor the equation
Next, we factor the left-hand side:
\[ x(x + 1) = 0 \]

### Step 5: Solve for \( x \)
Setting each factor equal to zero gives us:
1. \( x = 0 \)
2. \( x + 1 = 0 \) which simplifies to \( x = -1 \)

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