Problem 59 - Thomas Calculus in SI Units

Question

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2 / 3}(x+2)$$

Step-by-Step Solution

Answer

Critical point at \( x = -\frac{4}{5} \). Domain: all real numbers. Evaluate extreme points further if necessary.

Step 1: Find the Critical Points

To find the critical points, we first need the derivative of the function. Let \[ y = x^{2/3}(x+2). \]Apply the product rule: if \[ u = x^{2/3} \text{ and } v = x + 2, \]then \[ \frac{dy}{dx} = u'v + uv'. \]Calculate \[ u' = \frac{2}{3}x^{-1/3} \text{ and } v' = 1. \]Therefore, \[ \frac{dy}{dx} = \left(\frac{2}{3}x^{-1/3}(x+2)\right) + \left(x^{2/3}(1)\right). \]Combine terms:\[ \frac{dy}{dx} = \frac{2(x+2)}{3x^{1/3}} + x^{2/3}. \]Set this equation to zero to find critical points:\[ \frac{2(x+2)}{3x^{1/3}} + x^{2/3} = 0. \]

Step 2: Solve the Derivative Equation

To solve \[ \frac{2(x+2)}{3x^{1/3}} + x^{2/3} = 0, \]we multiply each term by \( 3x^{1/3} \) to clear the fraction:\[ 2(x+2) + 3x^{1/3}x^{2/3} = 0, \]which simplifies to:\[ 2x + 4 + 3x = 0. \]Combine like terms:\[ 5x + 4 = 0. \]Solving for \( x \), we get:\[ x = -\frac{4}{5}. \]

Step 3: Determine the Domain of the Function

The function \( y = x^{2/3}(x+2) \) is defined for all real \( x \), so its domain is \((-\infty, \infty)\).

Step 4: Evaluate the Endpoints and Critical Points

Evaluate the function at the critical point \( x = -\frac{4}{5} \).\[ y = \left(-\frac{4}{5}\right)^{2/3}\left(-\frac{4}{5}+2\right). \]Calculate:\[ y = \left(-\frac{4}{5}\right)^{2/3} \cdot \left(\frac{6}{5}\right). \]

Step 5: Identify Absolute and Local Extremes

Since the function is polynomial and defined over its entire range, consider values as \( x \to \pm \infty \).The function peaks or troughs at the critical point \( x = -\frac{4}{5} \).Evaluate further nearby or deeper calculations if necessary for precise values.

Key Concepts

Derivative

Derivatives are incredibly useful for understanding the behavior of functions. When you want to find how a function changes as its input changes, you look at its derivative. For the function \( y = x^{2/3}(x+2) \), the derivative tells us how this function changes at any point \( x \). The derivative is essentially the slope at any given point on the curve of the function.
Finding a derivative involves taking the original function and calculating the rate at which it changes. This usually requires some calculus rules.

• The **Power Rule** helps differentiate powers of \( x \), such as \( x^{2/3} \).
• The **Product Rule**, which we'll discuss next, is key when the function is a product of two simpler functions.

Product Rule

The product rule is a special rule in calculus that is used to find the derivative of a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), and you want to differentiate their product \( u(x) \, v(x) \), the product rule states:

• The derivative of \( u(x) \, v(x) \) is \( u'(x)\, v(x) + u(x) \, v'(x) \).

For our function \( y = x^{2/3}(x+2) \), we assign \( u = x^{2/3} \) and \( v = x + 2 \). By taking the derivative of each, \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 1 \), we apply the product rule:

• Combine \( u'v \) and \( uv' \) to get the complete derivative \( \frac{dy}{dx} = \frac{2(x+2)}{3x^{1/3}} + x^{2/3} \).

This equation helps us find critical points which are crucial for analyzing the function.

Extreme Values

Extreme values, specifically absolute and local extrema, are the highest or lowest points on a function. They indicate where a function reaches its maximum or minimum values. These points are found by determining critical points. Critical points occur where the derivative is zero or undefined.
In the example function \( y = x^{2/3}(x+2) \), once the derivative equation is set to zero, we solve it to find potential extrema. In this case, we find one critical point at \( x = -\frac{4}{5} \). Evaluating the function at this point and interpreting behavior as \( x \to \infty \) or \( x \to -\infty \) helps determine whether these are indeed local or absolute extrema.

• **Local Extremum**: A peak or trough in a smaller portion of the domain.
• **Absolute Extremum**: The highest or lowest value over the entire domain.

Domain of a Function

The domain of a function is the complete set of possible values of \( x \) which make the function real and defined. It tells us the 'allowed' values we can substitute into the function without creating undefined situations, such as division by zero or taking square roots of negative numbers.
For the function \( y = x^{2/3}(x+2) \), it is defined for all real numbers. This means the domain is \((-\infty, \infty)\). This particular function does not have any restrictions, such as fractions with a variable in the denominator or even roots.

• **Common domain issues** to watch for include divisions by zero and negative square roots.
• Since the function uses non-integer exponents, verification might be necessary, yet \( x^{2/3} \) is still defined for all real \( x \).