Problem 52 - Thomas Calculus

Question

In Exercises \(51-58\) , find the critical points and domain endpoints for each function. Then find the value of the function at each of these points and identify extreme values (absolute and local). $$ y=x^{2 / 3}\left(x^{2}-4\right) $$

Step-by-Step Solution

Answer

Critical points and local minima occur at \( x = 1 \), \( x = -1 \) with \( y = -3 \).

Step 1: Find the Domain of the Function

The given function is \( y = x^{2/3}(x^2 - 4) \). First, let's determine the domain of this function. The term \( x^{2/3} \) implies the function is defined for all real numbers \( x \) because the cube root is defined for all real numbers. Therefore, the domain is all real numbers \( \mathbb{R} \).

Step 2: Find the Derivative

To find the critical points, we first need to find the derivative of the function. The function is \( y = x^{2/3}(x^2 - 4) \). Use product rule: let \( u = x^{2/3} \) and \( v = x^2 - 4 \). Thus, \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 2x \). Then, \ \( y' = u'v + uv' = \frac{2}{3}x^{-1/3}(x^2 - 4) + x^{2/3}(2x) \).

Step 3: Simplify the Derivative Expression

Simplify the expression obtained from the derivative: \ \[ y' = \frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3} \]. \ This simplifies to: \ \[ y' = \frac{2(x^2 - 4)}{3x^{1/3}} + 2x^{5/3} \].

Step 4: Find Critical Points

Set the derivative equal to zero to find the critical points: \ \[ \frac{2(x^2 - 4)}{3x^{1/3}} + 2x^{5/3} = 0 \] \ Multiply through by \(3x^{1/3}\) to clear the fraction: \ \( 2(x^2 - 4) + 6x^{2} = 0 \) which simplifies to \( 8x^2 - 8 = 0 \). Solving this gives \( x^2 = 1 \), so the critical points are \( x = 1 \) and \( x = -1 \).

Step 5: Evaluate the Function at Critical Points and Domain Endpoints

The domain is all \( \mathbb{R} \). However, there are no finite endpoints since the domain is not bounded. Evaluate the function at the critical points: \ For \( x = 1 \), \( y = 1^{2/3}(1^2 - 4) = -3 \). \ For \( x = -1 \), \( y = (-1)^{2/3}((-1)^2 - 4) = -3 \).

Step 6: Identify Extreme Values

Evaluate the function as \( x \to \infty \) and \( x \to -\infty \): \ As \( x \to \infty \), \( y \to \infty \), and as \( x \to -\infty \), \( y \to -\infty \). \ Since both critical points give \( y = -3 \), these are identified as local minima.

Key Concepts

Function Domain

When determining the domain of a function, we focus on identifying which values of the variable make the function valid. For the function given, which is \( y = x^{2/3}(x^2 - 4) \), we find the domain by examining each term. The term \( x^{2/3} \) involves a cube root, which is defined for all real numbers \( x \). This means there are no restrictions from the cube root that would affect the domain. Additionally, \( x^2 - 4 \) is a simple polynomial, well-defined for all real \( x \). Thus, the domain of this function is all real numbers, represented by \( \mathbb{R} \). This provides a helpful foundation before analyzing the behavior of the function for any possible extreme values.

Derivative

The derivative is essential in finding critical points, which will help us identify extreme values. For a function like \( y = x^{2/3}(x^2 - 4) \), computing the derivative involves applying the product rule. This rule states that \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \). Here, we define \( u = x^{2/3} \) and \( v = x^2 - 4 \), leading to \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 2x \). Using these, the derivative is:

• \( y' = \frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3} \)

Simplifying this equation gives insight into where \( y' = 0 \) or where it is undefined, identifying the critical points and further helping us analyze local and extreme values.

Extreme Values

Extreme values refer to the maximum and minimum points on a function, either globally or locally. To find these points, we examine where the derivative is zero or undefined. For the function \( y = x^{2/3}(x^2 - 4) \), setting the derivative equal to zero gives us critical points:

• \( \frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3} = 0 \)
• This simplifies to \( 8x^2 - 8 = 0 \), which gives \( x = \pm 1 \)

Evaluating the function \( y \) at these points:

• For \( x = 1 \), \( y = -3 \)
• For \( x = -1 \), \( y = -3 \)

These values indicate the presence of local minima since the function value is the same for both points.

Local Minima

Local minima occur at points where the function value is less than all surrounding values, at least within some small interval. After evaluating our original function \( y = x^{2/3}(x^2 - 4) \) at the critical points \( x = 1 \) and \( x = -1 \), we find that the function equals \(-3\) at both. As neither point goes lower than \(-3\) in its immediate vicinity, these are identified as local minima. Even though \( y \to \infty \) as \( x \to \infty \), and \( y \to -\infty \) as \( x \to -\infty \), the critical points provide a sense of where the function dips before rising or falling as \( x \to \pm \infty \). This understanding is crucial for sketching the function and understanding its general behavior.