Problem 57 - University Calculus: Early Transcendentals

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

Local minima are at \( x = -1 \) and \( x = 1 \) with value 0; no maximum exists.

Step 1: Identify the Domain

The given function is \( y = \sqrt{x^2 - 1} \). The expression \( x^2 - 1 \) inside the square root must be non-negative. So, \( x^2 - 1 \geq 0 \), which implies \( x^2 \geq 1 \). This means \( x \leq -1 \) or \( x \geq 1 \). Therefore, the domain of the function is \( x \in (-\infty, -1] \cup [1, \infty) \).

Step 2: Find the Critical Points

To find the critical points, we need to differentiate the function. Let \( y = \sqrt{x^2 - 1} \), then the derivative is \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 1}} \). Critical points occur where the derivative is zero or undefined.\( \frac{x}{\sqrt{x^2 - 1}} = 0 \) implies that \( x = 0 \). However, \( x = 0 \) is not in the domain. Examine where the derivative is undefined: the derivative is undefined when \( \sqrt{x^2 - 1} = 0 \); hence for \( x = -1 \) and \( x = 1 \). These points are considered as critical points because the derivative changes from defined to undefined.

Step 3: Evaluate the Function at Critical and End Points

We determine the function values at the critical points and the bounds of the domain.- At \( x = -1 \), \( y = \sqrt{(-1)^2 - 1} = \sqrt{0} = 0 \).- At \( x = 1 \), \( y = \sqrt{1^2 - 1} = \sqrt{0} = 0 \). These calculations suggest that these points are potential local minima.

Step 4: Analyze Function Behavior at Domain Extrema

Consider the behavior as \( x \rightarrow -\infty \) and \( x \rightarrow \infty \). As \( x \) becomes increasingly negative or positive, \( \sqrt{x^2 - 1} \approx |x| \), which implies that the function value increases indefinitely in either direction. Thus, the function has no absolute maximum.

Step 5: Identify Local and Absolute Extrema

Based on the evaluation, there are local minima at \( x = -1 \) and \( x = 1 \) with corresponding function values of 0. There is no local or absolute maximum because the function increases without bound as \( x \to -\infty \) or \( x \to \infty \).

Key Concepts

Critical Points

In the quest to find extreme values of a function, determining its critical points is a crucial step. Critical points are where the function's derivative is either zero or undefined. These points allow us to uncover potential locations of local maxima or minima.
To find critical points for the function \( y = \sqrt{x^2 - 1} \), we first compute its derivative:

• The derivative is \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 1}} \).
• A critical point occurs when this derivative equals zero or becomes undefined.

If we set \( \frac{x}{\sqrt{x^2 - 1}} = 0 \), we find \( x = 0 \). However, \( x = 0 \) is not within the function's domain, so it is irrelevant for our purposes.
Instead, when the derivative is undefined (when the denominator \( \sqrt{x^2 - 1} \) is zero), it happens at the points \( x = -1 \) and \( x = 1 \). These are the valid critical points within our function's accepted domain.

Domain of a Function

Understanding the domain of a function is fundamental. It tells us where the function is defined and where to look for its critical points and extreme values.
For the function \( y = \sqrt{x^2 - 1} \), the domain consists of all \( x \) values that satisfy \( x^2 - 1 \geq 0 \). This condition ensures that the expression inside the square root is not negative, which would make the function imaginary.
By solving \( x^2 - 1 \geq 0 \), we find:

• \( x^2 \geq 1 \)
• This translates to \( x \leq -1 \) or \( x \geq 1 \).

Thus, the domain of \( y \) is \( x \in (-\infty, -1] \cup [1, \infty) \). This constraint restricts where we can analyze the function for extremum points.

Local Minima

Local minima are points where a function takes a smaller value than at nearby points. They are important in understanding a function's behavior. To identify local minima, we evaluate the function at critical points within its domain.

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

Both critical points, \( x = -1 \) and \( x = 1 \), provide a function value of zero, indicating these are local minima.
No local maximum exists since the function increases without bound as \( x \to -\infty \) or \( x \to \infty \). This behavior reflects in the function's open-ended nature.

Differentiation

Differentiation is the mathematical process of finding a function's derivative. This tool helps us determine rate changes and identify critical points for extreme values.
The derivative of \( y = \sqrt{x^2 - 1} \) is found using:

• The chain rule: \( \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 1}} \).

Differentiation uncovers where the function's slope is zero or undefined, leading us to critical points.
Calculating the derivative equips us with insights about a function's behavior. This understanding of increasing or decreasing tendencies helps identify potential extrema. As in this problem, noticing where the derivative is zero or undefined pinpoints key features of the function.