Problem 40 - Thomas Calculus

Question

In Exercises \(35-44,\) find the extreme values of the function and where they occur. $$ y=\frac{1}{\sqrt{1-x^{2}}} $$

Step-by-Step Solution

Answer

The function has a local minimum at \( x = 0 \). No finite extrema occur at the boundaries.

Step 1: Find the Domain of the Function

The function is given by \( y = \frac{1}{\sqrt{1-x^2}} \). The expression inside the square root \( 1-x^2 \) must be greater than zero, meaning \( x^2 < 1 \). Therefore, the domain of the function is \(-1 < x < 1\).

Step 2: Find the First Derivative

To determine extreme values, we first need to find the derivative of the function. Let \( f(x) = \frac{1}{\sqrt{1-x^2}} \). Using chain rule, the derivative is:\[ f'(x) = \frac{d}{dx} \left( 1-x^2 \right)^{-1/2} = -\frac{1}{2} (1-x^2)^{-3/2} \cdot (-2x) = \frac{x}{(1-x^2)^{3/2}}. \]

Step 3: Set the First Derivative to Zero and Solve for x

To find critical points, set \( \frac{x}{(1-x^2)^{3/2}} = 0 \). This implies:\[ x = 0. \]We note that at \( x = 0 \), the first derivative is zero, hence it is a critical point.

Step 4: Determine the Nature of the Critical Point

To confirm whether the critical point at \( x = 0 \) is a maximum, minimum, or neither, use the second derivative test. Compute the second derivative:\[ f''(x) = \frac{(1-x^2)^3/2 - x(1-x^2)^{1/2} \cdot -3x}{(1-x^2)^3} = \frac{1-2x^2}{(1-x^2)^{5/2}}. \]Evaluate the second derivative at \( x = 0 \):\[ f''(0) = \frac{1}{1^2} = 1. \]Since \( f''(0) > 0 \), \( x = 0 \) is a local minimum.

Step 5: Verify Extrema at Domain Boundaries

Given the domain is \(-1 < x < 1\), check the behavior of the function as \( x \) approaches \( -1 \) and \( 1 \). As \( x \to \pm 1 \), \( \sqrt{1-x^2} \to 0 \) causing \( y \to \infty \). This implies there are no finite maximum or minimum values at the boundaries, confirming the initial findings.

Key Concepts

Domain of the Function

The domain of a function refers to all the possible input values (usually represented by \( x \)) for which the function is defined. For the function \( y = \frac{1}{\sqrt{1-x^2}} \), we need to ensure we only include values for \( x \) that don't lead to mathematical mishaps, like division by zero or taking the square root of a negative number. In this case, \( 1-x^2 \) must be greater than zero so that we can take its square root.
To satisfy this condition:

• The expression \( 1-x^2 > 0 \) translates to \( x^2 < 1 \).
• This implies \(-1 < x < 1\).

Thus, the domain of the function is all \( x \) values between \(-1\) and \(1\), not including \(-1\) and \(1\) themselves. This ensures the function remains valid and calculable.

First Derivative

The first derivative of a function, denoted as \( f'(x) \), helps us find the rate of change of the function. It's crucial for identifying the points where the function has horizontal tangents, known as critical points. For the function \( y = \frac{1}{\sqrt{1-x^2}} \), applying the chain rule results in the following derivative:\[ f'(x) = \frac{x}{(1-x^2)^{3/2}}.\]This first derivative expression tells us how \( y \) changes as \( x \) varies, especially how its slope approaches zero.

• It simplifies the task of locating critical points, which are candidates for extreme values like maxima or minima.

In this case, understanding \( f'(x) \) is a stepping stone to further investigate the function's behavior at various \( x \) values.

Second Derivative Test

Once we find the critical points using the first derivative, the second derivative test comes in handy to determine their nature. The second derivative, \( f''(x) \), provides information about the concavity of the function. If it's concave up at a point, it's a local minimum; if concave down, it's a local maximum.
For our function, the second derivative is:\[ f''(x) = \frac{1-2x^2}{(1-x^2)^{5/2}}.\]Evaluating this at \( x = 0 \) gives\[ f''(0) = 1,\]indicating concave upwards, confirming a local minimum at \( x=0 \).

• This test is essential for validation as it ensures the accuracy of identified extrema.

Thus, it helps conclude the nature of the critical points in our analysis.

Critical Points

Critical points are key in analyzing a function, as they indicate where the function's slope is zero or the derivative is undefined. These points are where the function might attain its extreme values, like peaks or valleys. For \( y = \frac{1}{\sqrt{1-x^2}} \), setting the first derivative to zero reveals the critical point:\[ \frac{x}{(1-x^2)^{3/2}} = 0 \]leads to \( x = 0 \).

• This means when \( x = 0 \), the function might reach either a maximum or a minimum.

After applying the second derivative test, we confirm \( x = 0 \) is a local minimum. Understanding critical points allows us to deeply comprehend how the function behaves at various inputs within the domain.