Problem 53 - Thomas Calculus in SI Units

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=\sqrt{x^{2}-1}$$

Step-by-Step Solution

Answer

Absolute minima at \( x = \pm 1 \). No local maxima or minima elsewhere.

Step 1: Identify the Domain

The function is given by \( y = \sqrt{x^{2} - 1} \). The expression inside the square root must be non-negative. Therefore, the domain is determined by the inequality \( x^2 - 1 \geq 0 \). Solving this inequality gives \( x \leq -1 \) or \( x \geq 1 \). Thus, the domain of \( y \) is \( (-\infty, -1] \cup [1, \infty) \).

Step 2: Find Critical Points

To find critical points, we differentiate the function. The derivative of \( y \) with respect to \( x \) is derived from the chain rule. \( y = \sqrt{x^2 - 1} \) implies \( y' = \frac{x}{\sqrt{x^2 - 1}} \). Critical points occur where \( y' = 0 \) or \( y' \) is undefined. Here, \( y' = 0 \) when \( x = 0 \), which is not in the domain of \( y \). The derivative is undefined where the denominator is zero, which happens when \( x = \pm 1 \). Both are in the domain.

Step 3: Evaluate Function at Endpoints and Critical Points

Evaluate \( y \) at the endpoints and critical points in the domain. For the endpoint \( x = 1 \), \( y = \sqrt{1^2 - 1} = 0 \) and for \( x = -1 \), \( y = \sqrt{(-1)^2 - 1} = 0 \). At the critical points, though the domain does not technically include \( x = 0 \), \( y \) is zero at both endpoints, confirming \( (x, y) = (1,0) \) and \( (x, y) = (-1,0) \).

Step 4: Determine the Nature of Extremes

The second derivative is not needed here; using domain limits, observe that as \( x \to \pm \infty \), \( y \to \infty \). Therefore, at both endpoints \( x = 1 \) and \( x = -1 \), \( y = 0 \) represents absolute minima. There are no local maxima or minima between these points as the curve trends upwards towards infinity, without turning back.

Key Concepts

Domain of a Function

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

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

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.