Problem 37 - Calculus of a Single Variable

Question

Using the Second Derivative Test In Exercises \(31-42\) , find all relative extrema. Use the Second Derivative Test where applicable. $$ f(x)=x^{2 / 3}-3 $$

Step-by-Step Solution

Answer

The Second Derivative Test was inconclusive at \(x=0\). Additional analysis with other methods is needed to determine if this point is a maximum, minimum, or neither.

Step 1: Find the Derivative

The first step is to find the derivative of the function using the power rule for derivatives. The derivative \(f'(x)=\frac{2}{3}x^{-1/3}\).

Step 2: Find the Critical Points

Next, set the derivative equal to zero and solve for \(x\) to find the critical points. The solution is \(x=0\).

Step 3: Apply the Second Derivative Test

Find the second derivative using the chain rule. It's \(f''(x)=-\frac{2}{9}x^{-4/3}\). To apply the Second Derivative Test, evaluate the second derivative at \(x=0\). However, at \(x=0\), the second derivative is undefined, which means the Second Derivative Test is inconclusive. This is a drawback of the Second Derivative Test. In cases where it's inconclusive, usually the First Derivative Test or other methods can be used instead to understand the function's behavior at that point.

Key Concepts

Critical Points in Calculus

Understanding the concept of critical points is fundamental for solving various calculus problems. Critical points are the 'x' values where the first derivative of a function is either zero or undefined. Locating these points is essential when analyzing the behavior of functions, especially when identifying local extreme values—that is, the high and low points on a graph.

Usually, the process involves taking the derivative of a function and setting it equal to zero. Take, for example, the function f(x) = x^{2/3} - 3. The first derivative, found by applying the power rule, is f'(x) = \( \frac{2}{3}x^{-1/3} \). By setting this derivative to zero, we find that x = 0 is the critical point for this function.

However, it's important not just to find the critical points but to determine what they represent. Are they maximums, minimums, or points of inflection? This is where various tests, like the First Derivative Test or the Second Derivative Test, come in to provide more information. In our example, since the second derivative is undefined at x=0, the Second Derivative Test is inconclusive, prompting the need for alternative analysis methods.

Power Rule for Derivatives

A core concept in calculus is taking derivatives, and the power rule offers a direct method for differentiating functions with power terms. Here's the power rule at a glance: if f(x) = x^n, then f'(x) = nx^{n-1}, where n is any real number. It simplifies finding the slope of the curve at any point.

Looking at our exercise function f(x) = x^{2/3}-3, we use the power rule to get its derivative. Applying f'(x) = \( \frac{2}{3}x^{-1/3} \). This simplicity is why the power rule is a favorite tool in calculus, efficient and applicable to a wide range of functions. Yet, it's just as important to remember its limitations - the power rule alone can't handle more complex functions involving products, quotients, or compositions of functions.

Chain Rule for Derivatives

When functions become more complex, involving compositions of functions (that is, functions within functions), the chain rule is a crucial derivative tool that comes to play. The chain rule states that if you have a composite function f(g(x)), the derivative is the product of the derivative of f with respect to g, and the derivative of g with respect to x, or mathematically: f'(x) = f'(g(x)) · g'(x).

Returning to our sample exercise, for the second derivative of f(x), we'd need the chain rule if f(x) was a composition of functions. Here, however, the second derivative f''(x) = -\( \frac{2}{9}x^{-4/3} \) is still calculated directly since there isn't an outer function enveloping the x^{2/3} term. Still, the chain rule is invaluable for many functions that students will encounter, including trigonometric, exponential, and logarithmic functions, and mastering it is key for successful calculus problem-solving.