Consider the function `f(x) = (x^(2)-1), where `x in R`
At what value of x does f(x) attain minimum value?

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find the value of \( x \) at which the function \( f(x) = x^2 - 1 \) attains its minimum value, we will follow these steps:

### Step 1: Find the first derivative of the function
The first step is to differentiate the function \( f(x) \).

\[ f(x) = x^2 - 1 \]

Taking the derivative with respect to \( x \):

\[ f'(x) = \frac{d}{dx}(x^2 - 1) = 2x \]

### Step 2: Set the first derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ f'(x) = 2x = 0 \]

Solving for \( x \):

\[ x = 0 \]

### Step 3: Find the second derivative
Next, we need to determine whether this critical point is a minimum or maximum by finding the second derivative of the function.

\[ f''(x) = \frac{d}{dx}(2x) = 2 \]

### Step 4: Analyze the second derivative
Now we evaluate the second derivative at the critical point \( x = 0 \):

\[ f''(0) = 2 \]

Since \( f''(0) > 0 \), this indicates that the function is concave up at \( x = 0 \), which means that \( x = 0 \) is a point of minimum.

### Conclusion
Thus, the function \( f(x) = x^2 - 1 \) attains its minimum value at:

\[ \boxed{0} \]