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Critical points are where a function's derivative is either zero or undefined. For the function \( f(x) = \frac{x^2 - 1}{x^2 + 1} \), critical points help us determine where the function might have a maximum, minimum, or plateau. Finding these points starts with taking the derivative of the function.
For \( f(x) \), the derivative \( f'(x) = \frac{4x}{(x^2+1)^2} \). To find the critical points, we set \( f'(x) = 0 \). Here, the numerator must be zero because the denominator can't be zero (as it would make the expression undefined).
Thus, the critical point is obtained when \( x = 0 \). At this point, we can then evaluate the function to understand the nature of \( f(x) \) at this critical point. Plugging \( x = 0 \) back into the function gives us \( f(0) = -1 \), indicating a potential minimum value.

The derivative test is a method to determine if a critical point is a maximum, minimum, or neither. For our function, we use the first derivative test. This involves analyzing the sign of \( f'(x) \) around the critical point found.
When \( x < 0 \), \( f'(x) = \frac{4x}{(x^2+1)^2} \) results in a negative value, indicating that the function is decreasing. Conversely, when \( x > 0 \), \( f'(x) \) is positive, showing the function is increasing. The change from negative to positive across \( x = 0 \) suggests that this critical point is indeed a minimum.
This implies that at \( x = 0 \), the function transitions from decreasing to increasing, confirming that \( f(0) = -1 \) is the minimum value of the function.

The behavior of a function as \( x \) approaches infinity, both positively and negatively, helps us understand its limits and range. For \( f(x) = \frac{x^2 - 1}{x^2 + 1} \), evaluating how the function behaves when \( x \) tends towards infinity or negative infinity confirms the range of possible values that \( f(x) \) can take.
As \( x \rightarrow +\infty \) or \( x \rightarrow -\infty \), notice that both \( x^2 - 1 \) and \( x^2 + 1 \) are dominated by the \( x^2 \) term, making the fraction approximate to \( \frac{x^2}{x^2} = 1 \). However, the function approaches this limit from just below, so \( f(x) \rightarrow 1^- \).
This behavior confirms that while the function approaches 1 as \( x \) travels to infinity, it never actually reaches this value, thus helping establish the range of \( (-1, 1) \) for \( f(x) \). Together, with the critical point evaluation, this shows that the minimum value the function can attain is \(-1\).