If `f(x)=x^(2)+2x-2` and if `f(s-1)=1`, what is the smallest possible value of s ?

Options

A. `-3`
B. `-2`
C. `-1`
D. 1

Explanation

To solve the problem, we will follow these steps:

1. **Substitute \( s - 1 \) into the function \( f(x) \)**:
We know that \( f(x) = x^2 + 2x - 2 \). We need to find \( f(s - 1) \):
\[ f(s - 1) = (s - 1)^2 + 2(s - 1) - 2 \]

2. **Set the equation equal to 1**:
According to the problem, \( f(s - 1) = 1 \):
\[ (s - 1)^2 + 2(s - 1) - 2 = 1 \]

3. **Simplify the equation**:
First, expand \( (s - 1)^2 \):
\[ (s - 1)^2 = s^2 - 2s + 1 \]
Now substitute this back into the equation:
\[ s^2 - 2s + 1 + 2(s - 1) - 2 = 1 \]
Simplifying further:
\[ s^2 - 2s + 1 + 2s - 2 - 2 = 1 \]
Combine like terms:
\[ s^2 - 3 = 1 \]

4. **Rearranging the equation**:
Add 3 to both sides:
\[ s^2 = 4 \]

5. **Taking the square root**:
Taking the square root of both sides gives us:
\[ s = \pm 2 \]

6. **Finding the smallest possible value of \( s \)**:
The two possible values for \( s \) are \( 2 \) and \( -2 \). The smallest possible value is:
\[ s = -2 \]

Thus, the smallest possible value of \( s \) is \( -2 \).