Problem 51 - 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). $$ y=x^{2 / 3}(x+2) $$

Step-by-Step Solution

Answer

Critical points: \( x = -1, 0 \). Local maximum at \( y = 1 \) when \( x = -1 \), local minimum at \( y = 0 \) when \( x = 0 \).

Step 1: Identify the function

The given function is \( y = x^{2/3}(x + 2) \). We are tasked to find the critical points and domain endpoints.

Step 2: Determine the domain

Since the function involves \( x^{2/3} \), which is defined for all real \( x \), the domain of the function is all real numbers, \( (-\infty, \infty) \).

Step 3: Find the derivative

Use the product rule to differentiate the function: \( y = u \cdot v \) where \( u = x^{2/3} \) and \( v = x + 2 \). First, find \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 1 \).Then, the derivative \( y' = u'v + uv' = \frac{2}{3}x^{-1/3}(x + 2) + x^{2/3} \cdot 1 \).Simplify the expression to \( y' = \frac{2(x + 2)}{3x^{1/3}} + x^{2/3} \).

Step 4: Find critical points

Critical points occur where the derivative is zero or undefined. Set \( y' = 0 \), thus \( \frac{2(x + 2)}{3x^{1/3}} + x^{2/3}= 0 \). Simplifying, yields critical points by solving the equation. The derivative is also undefined at \( x = 0 \). Thus, potential critical points are \( x = -1, 0 \).

Step 5: Evaluate function at critical points

Substitute the critical points back into \( y = x^{2/3}(x+2) \): - At \( x = -1 \), \( y = (-1)^{2/3} \cdot 1 = 1 \).- At \( x = 0 \), \( y = 0^{2/3} \cdot 2 = 0 \).

Step 6: Evaluate the function at endpoints of the domain

Since the domain is \((-\infty, \infty)\), there are no finite endpoints to evaluate. However, examine the limits: - As \( x \to \infty \), \( y \to \infty \). - As \( x \to -\infty \), since \( x+2 \to -\infty \), \( y \to -\infty \).

Step 7: Identify extreme values

From the evaluations, \( x = -1 \) gives \( y = 1 \) and is a local maximum since the value goes to infinity at both ends. At \( x = 0 \), \( y = 0 \) might be considered a local minimum as it is the lowest point compared to the surrounding points close to zero.

Key Concepts

Derivative

A derivative measures how a function changes as its input changes. Here, the function given is \( y = x^{2/3}(x + 2) \). To find the derivative, we apply the **product rule**, which is useful when differentiating a product of two functions. The product rule states: for functions \( u \) and \( v \), the derivative of their product \( y = uv \) is \( y' = u'v + uv' \). Let's break it down step-by-step:

• Define \( u = x^{2/3} \), with the derivative \( u' = \frac{2}{3}x^{-1/3} \).
• Define \( v = x + 2 \), with the derivative \( v' = 1 \).

Now apply the product rule:\[ y' = \left( \frac{2}{3}x^{-1/3} \right)(x + 2) + x^{2/3} \cdot 1 \]This helps in finding points where the function might have a local increase or decrease, known as critical points.

Domain

The domain of a function is all the inputs (or \( x \)-values) for which the function is defined. For the function \( y = x^{2/3}(x + 2) \), the base expression involved is \( x^{2/3} \). The root \( x^{1/3} \) is defined for all real numbers, meaning there are no restrictions on the values \( x \) can take. Therefore, for this function, the domain is \((-\infty, \infty) \). This means you can input any real number into the function, and you'll get an output. Knowing the domain is crucial, as it tells us the range of \( x \) that we should consider when evaluating extreme values.

Extreme Values

Extreme values of a function are points at which the function reaches maximum or minimum values. These can be local (compared to points close by) or absolute (the highest or lowest in the entire domain). For the given function, potential extreme values occur at critical points where the derivative is zero or doesn't exist (undefined). The critical points calculated were at \( x = -1 \) and \( x = 0 \):

• At \( x = -1 \), substituting, gives \( y = 1 \). This is a local maximum because the function increases to \( \infty \) for higher \( x \) and decreases to \(-\infty \) for lower \( x \).
• At \( x = 0 \), substituting, gives \( y = 0 \). This is a local minimum since any nearby function value is higher than zero.

Understanding these critical and extreme points is incredibly helpful in sketching the graph and understanding the function's behavior.

Product Rule

The product rule is a handy derivative rule used when differentiating products of two functions, ensuring that each part contributes correctly to the result. The formula for the product rule is:\[ (uv)' = u'v + uv' \]where \( u \) and \( v \) are functions of \( x \). In practical terms, to apply:

• Differentiating \( u \), compute \( u' \).
• Leave \( v \) as it is and multiply it by \( u' \).
• Then, differentiate \( v \) to find \( v' \).
• Multiply \( u \) by \( v' \).
• Finally, add both resulting products together to get \( y' \).

In the exercise, using the product rule allowed us to accurately find the function's derivative, essential for solving for critical points and analyzing the function's behavior.