Consider the function `f(X)=(x^(2)-1)/(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

Explanation

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

### Step 1: Find the derivative of \( f(x) \)

We start by applying the quotient rule to find the derivative \( f'(x) \):

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

### Step 2: Simplify the derivative

Now, we simplify the expression:

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

This simplifies to:

\[ f'(x) = \frac{2x(2)}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2} \]

### Step 3: Set the derivative equal to zero

To find the critical points, we set the derivative equal to zero:

\[ \frac{4x}{(x^2 + 1)^2} = 0 \]

This gives us:

\[ 4x = 0 \implies x = 0 \]

### Step 4: Determine the nature of the critical point

Next, we need to check whether this critical point is a minimum or maximum. We can do this by finding the second derivative \( f''(x) \) or by using the first derivative test.

Let’s find the second derivative:

1. Differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx} \left( \frac{4x}{(x^2 + 1)^2} \right) \]

Using the quotient rule again:

\[ f''(x) = \frac{(x^2 + 1)^2(4) - 4x(2)(x)(x^2 + 1)}{(x^2 + 1)^4} \]

2. Simplifying \( f''(x) \):

After simplification, we can evaluate \( f''(0) \):

\[ f''(0) = \frac{(0^2 + 1)^2(4) - 4(0)(2)(0)(0^2 + 1)}{(0^2 + 1)^4} = \frac{4}{1} = 4 \]

Since \( f''(0) > 0 \), this indicates that \( x = 0 \) is a point of local minimum.

### Step 5: Find the minimum value of \( f(x) \)

Now we can find the minimum value of \( f(x) \) at \( x = 0 \):

\[ f(0) = \frac{0^2 - 1}{0^2 + 1} = \frac{-1}{1} = -1 \]

### Conclusion

Thus, the function \( f(x) \) attains its minimum value at \( x = 0 \) and the minimum value is \( -1 \).

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