Problem 55 - Thomas Calculus in SI Units

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=\frac{1}{\sqrt[3]{1-x^{2}}}$$

Step-by-Step Solution

Answer

Local and absolute minimum at \( x = 0 \); function goes to infinity near \( x = \pm 1 \).

Step 1: Define the Domain

The given function is \( y = \frac{1}{\sqrt[3]{1-x^2}} \). The expression inside the cube root, \(1 - x^2\), must be non-zero since division by zero is undefined. Hence, the domain is \( x \in (-1, 1) \) because \(1 - x^2\) is positive in this interval.

Step 2: Find the Derivative

To find extreme values, we'll need the first derivative. The function can be rewritten as \( y = (1-x^2)^{-1/3} \). Using the chain rule, the derivative is \( y' = \frac{d}{dx}(1-x^2)^{-1/3} = \frac{2x}{3}(1-x^2)^{-4/3} \).

Step 3: Find Critical Points

Set the derivative \( y' = \frac{2x}{3}(1-x^2)^{-4/3} = 0 \). This occurs when \( 2x = 0 \), thus \( x = 0 \). Since the cube root functions are defined on the entire real line, \( x = 0 \) is the only critical point.

Step 4: Determine the Nature of Critical Points

To determine if \( x = 0 \) is a maximum or minimum, evaluate the second derivative or consider the sign change of the first derivative around it. However, it's easier here to infer by testing values around \( x = 0 \). For instance, test \( y' \) at \( x = -0.5 \) and \( x = 0.5 \). Since \( y' > 0 \) on both sides of \( x = 0 \), \( x = 0 \) is a point where \( y \) steadies, indicating a local minimum.

Step 5: Check the Values at the Boundary

Examine the behavior of \( y \) as \( x \to 1^- \) and \( x \to -1^+ \). As \( x \to 1^- \) or \( x \to -1^+ \), \((1-x^2)^{-1/3} \to \infty \). Thus, \( y \) approaches infinity and there are no finite absolute maximum or minimum, only an absolute minimum at \( x = 0 \).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (usually "x" values) that a function can accept. In our specific example, the function is defined as \( y = \frac{1}{\sqrt[3]{1-x^2}} \). The key here is ensuring that the denominator is not zero. Since division by zero is undefined in mathematics, we must determine where the expression \( 1 - x^2 \) is non-zero. To do this, we solve the inequality \( 1 - x^2 eq 0 \).For this function:

• The expression \( 1 - x^2 \) is zero when \( x = \pm 1 \).
• Thus, the domain is the interval \( (-1, 1) \), where \( x \) can take any value except exactly -1 or 1.

This domain ensures the function remains defined, as when \( x \) approaches -1 or 1 from within the interval, the function remains valid and non-zero.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are crucial as potential locations of maxima, minima, or saddle points — in simpler terms, they are where the function might change direction.
For our function, after finding the derivative \( y' = \frac{2x}{3}(1-x^2)^{-4/3} \), we look for values of \( x \) where the derivative equals zero. Solving \( \frac{2x}{3}(1-x^2)^{-4/3} = 0 \) gives:

• \( 2x = 0 \) which means \( x = 0 \).

This means \( x = 0 \) is the potential critical point because the derivative changes, indicating a potential shift in the behavior of the function. There are no other critical points within the interval \(-1, 1)\), as the derivative does not equal zero or become undefined elsewhere within this range.

Derivative

A derivative tells us how a function changes at any given point – essentially, it gives us the slope of the function or the rate of change. For the function \( y = \frac{1}{\sqrt[3]{1-x^2}} \), we need to find its derivative to understand where the function might have a peak or valley.
Using the chain rule, the derivative is calculated as:\[ y' = \frac{2x}{3}(1-x^2)^{-4/3} \]This derivative:

• Is zero at \( x = 0 \), implying the possibility of an extreme value.
• Is never undefined within the domain \((-1, 1)\), as the denominator part \((1-x^2)^{-4/3}\) is fine as \( 1-x^2 \) never reaches zero within that domain.

By examining this derivative, we can make sound judgments about the behavior of the original function, particularly near the potential critical point \( x = 0 \).

Local Minimum

A local minimum is a point on the graph of a function where the function value is smaller than nearby values. It represents a "valley" in the function. To determine if a point is a local minimum, one could check the sign of the first derivative before and after the point or use the second derivative test.
In our function's case, \( x = 0 \) was determined to be a critical point. To confirm it is a local minimum, we check nearby values. Observing that the derivative \( y' = \frac{2x}{3}(1-x^2)^{-4/3} \) is:

• Positive before and after 0, meaning the function transitions from decreasing to increasing.

This observation confirms that at \( x = 0 \), the function indeed has a local minimum. The function does not dip below its value at \( x = 0 \) anywhere within its domain.