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Understanding the domain of a function is crucial, as it tells you all the possible input values (x-values) for which the function is defined. For the function \( y = \sqrt{x^{2} - 1} \), we need to look inside the square root. We know that we cannot take the square root of a negative number in the real number system.
This means the expression \( x^2 - 1 \) must be greater than or equal to zero.To find the domain of \( f(x) = \sqrt{x^{2} - 1} \), we set up the inequality: \[ x^2 - 1 \geq 0 \]Breaking it down leads to two possible inequalities:

• \( x^2 \geq 1 \)
• \( x \leq -1 \) or \( x \geq 1 \)

This gives us the domain as \( (-\infty, -1] \cup [1, \infty) \). It means that the function is defined for all x-values less than or equal to -1 or greater than or equal to 1. Values between -1 and 1 are excluded because they make the expression inside the square root negative.
Mastering domain identification is essential for understanding where a function operates.

Critical points of a function occur where the derivative is either zero or undefined. These points are significant because they are potential locations for local maxima, minima, or points of inflection.For the function \( y = \sqrt{x^2 - 1} \), we first differentiate using the chain rule:

• \( y' = \frac{x}{\sqrt{x^2 - 1}} \)

Critical points are found where \( y' = 0 \) or \( y' \) is undefined. A critical point occurs when the numerator in the derivative equals zero: \( x = 0 \).
However, \( x = 0 \) isn't in the domain, so we ignore it here.Instead, we look at where \( y' \) is undefined, which happens if the denominator equals zero:

• \( x = \pm 1 \)

These values are within the domain, making them critical points. Thus, our critical points occur at \( x = 1 \) and \( x = -1 \), corresponding to the endpoints of our domain.
Finding these points helps in analyzing the behavior of the function along its domain.

Absolute minima are the lowest points on a function over its entire domain. They are important because they represent the smallest value a function can reach.After finding the critical points of \( y = \sqrt{x^2 - 1} \), we evaluate the function at these points and the endpoints of the domain. Calculate as follows:

• At \( x = 1 \), \( y = \sqrt{1^2 - 1} = 0 \)
• At \( x = -1 \), \( y = \sqrt{(-1)^2 - 1} = 0 \)

Both points give a function value of zero. Since these are also the endpoints of the domain where the behavior of the graph changes, these are the absolute minima.
In this scenario, \( y = 0 \) at both \( x = 1 \) and \( x = -1 \) represents the minimum values throughout the entire domain. There are no absolute maxima in the domains \((-\infty, -1] \) and \([1, \infty) \) because the function approaches infinity as \( x \to \pm \infty \). Recognizing absolute minima thus clarifies the function's range and lowest values.