Problem 5 - Calculus of a Single Variable

Question

Finding the Value of the Derivative at Relative Extrema In Exercises \(1-6,\) find the value of the derivative (if it exists) at each indicated extremum. $$ f(x)=(x+2)^{2 / 3} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=(x+2)^{2 / 3}\), \(f'(x)\), does not exist at the only critical point \(x = -2\).

Step 1: Compute the derivative of the function

First we must calculate the derivative of the function \(f(x)\). To do this, we need to apply the chain rule. The first step is to bring the exponent in front and decrement the exponent by one and then differentiate the terms inside the bracket considering it as a composite function. Thus, \(f'(x) = \frac{2}{3} (x+2)^{-\frac{1}{3}} \cdot 1\), which simplifies to \(f'(x) = \frac{2}{3 (x+2)^{\frac{1}{3}}}\)

Step 2: Identify the Critical Points

Critical points occur where the derivative is zero or doesn't exist. For the function \(f'(x)\), the derivative doesn't exist for \(x = -2\) because it will lead to a division by zero. Thus, \(x = -2\) is the only critical point.

Step 3: Evaluate the Derivative at the Critical Points

Given that \(x = -2\) is a critical point, we can find the value of the derivative at \(x = -2\). However, as we determined in Step 2, the derivative does not exist at \(x = -2\). As such, there is no value for \(f'(-2)\).

Key Concepts

The Chain Rule

Understanding the chain rule in calculus is essential for computing the derivatives of composite functions. In essence, the chain rule provides a method to differentiate a function that comprises one or more functions. In simpler terms, if you have a 'function within a function', the chain rule is your go-to tool.
The chain rule states that if you have a composite function, say, \( g(f(x)) \), then the derivative of this composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Mathematically, this can be represented as \( g'(f(x)) \cdot f'(x) \).
To apply this to our original exercise, \( f(x) = (x+2)^{\frac{2}{3}} \), we treat \( (x+2) \) as the inner function and \( z^{\frac{2}{3}} \) as the outer function, where \( z = (x+2) \). According to the chain rule, we differentiate \( z^{\frac{2}{3}} \) with respect to \( z \) and \( (x+2) \) with respect to \( x \) and then multiply these derivatives together. To make the concept easier to comprehend, we often use simpler notation and algebraic manipulation for clarity.

Critical Points

Critical points are pivotal in the study of calculus; they are where the derivative of a function is either zero or undefined. These points are the locations on the graph of a function where the slope of the tangent line is horizontal (derivative is zero) or the tangent line is not defined (derivative doesn't exist).
Finding Critical Points:

• First, find the derivative of the function in question.
• Next, set the derivative equal to zero and solve for \( x \).
• Also, determine where the derivative doesn't exist.

For the function in our exercise, \( f'(x) = \frac{2}{3 (x+2)^{\frac{1}{3}}} \) provides a clue to a critical point. By analyzing where the derivative might not exist (division by zero in this case), we pinpoint \( x = -2 \) as a critical point because plugging this into the derivative results in an undefined expression.

Calculus

Calculus, the branch of mathematics that deals with derivatives and integrals, is ubiquitous in the world of scientific inquiry and problem-solving. It provides tools for modeling and solving problems involving change and space. In the context of our exercise, calculus allows us to explore and understand the behavior of the function \( f(x) = (x+2)^{\frac{2}{3}} \) through its derivative.
The act of finding the derivative, which involves computing the rate at which a function is changing at any given point, is a foundational concept in calculus. In this case, despite finding the derivative, we encounter an interesting situation at the critical point where the derivative doesn't exist. Such insights illustrate the nuanced nature of calculus where not only the existence of derivatives but also their behavior at different points becomes a gateway to comprehending deeper properties of functions.