Problem 12 - Essential Calculus Early Transcendentals

Question

Find the local maximum and minimum values of \(f\) using both the First and Second Derivative Tests. Which method do you prefer? \(f(x)=\frac{x^{2}}{x-1}\)

Step-by-Step Solution

Answer

Local max at \(x=2\), local min at \(x=0\). Preference: Second Derivative Test (simpler for confirming critical points).

Step 1: Find the First Derivative

To find the first derivative of the function \( f(x) = \frac{x^2}{x-1} \), we apply the quotient rule. The quotient rule states that if \( u = x^2 \) and \( v = x-1 \), then \( f' = \frac{u'v - uv'}{v^2} \). After differentiating, \( u' = 2x \) and \( v' = 1 \), so \( f'(x) = \frac{2x(x-1) - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2} \).

Step 2: Identify Critical Points using the First Derivative Test

To find critical points, set \( f'(x) = 0 \). Solving \( \frac{x(x-2)}{(x-1)^2} = 0 \) gives critical points \( x = 0 \) and \( x = 2 \). Since \( x - 1 \) in the denominator cannot be zero, \( x = 1 \) is a point of discontinuity.

Step 3: Determine the Nature of Critical Points using the First Derivative Test

Check intervals around the critical points. For \( x \in (-\infty, 0) \), the derivative is negative; for \( x \in (0, 1) \), it's positive. Hence, \( x = 0 \) is a local minimum. For \( x \in (1, 2) \), the derivative is positive, and it changes to negative for \( x \in (2, \infty) \), indicating \( x = 2 \) is a local maximum.

Step 4: Find the Second Derivative

Differentiate the first derivative \( f'(x) = \frac{x(x-2)}{(x-1)^2} \) using the quotient rule again. After simplification, the second derivative is \( f''(x) = \frac{2}{(x-1)^3} \).

Step 5: Confirm Critical Points using the Second Derivative Test

Evaluate the second derivative at critical points. For \( x = 0 \), \( f''(0) = \frac{2}{(-1)^3} = -2 \), indicating a local minimum. For \( x = 2 \), \( f''(2) = \frac{2}{1^3} = 2 \), indicating a local maximum.

Step 6: Evaluate and Compare Methods

Both methods identify \( x = 2 \) as a local maximum and \( x = 0 \) as a local minimum. The First Derivative Test involves analyzing intervals, while the Second Derivative Test needs only the sign of the second derivative at the critical points.

Key Concepts

First Derivative Test

The First Derivative Test is a useful tool in calculus to determine if critical points of a function are local maxima or minima. To apply this test, first find the derivative of the function and identify the critical points where the derivative is zero or undefined.

Once you have the critical points, examine the signs of the derivative before and after each point:

• If the derivative changes from negative to positive, the critical point is a local minimum.
• If it changes from positive to negative, it's a local maximum.
• If there's no change, the critical point is not a local extremum.

In the original exercise, the function is analyzed at different intervals based on the critical points: 0 and 2. The behavior of the function in these intervals helps determine that 0 is a local minimum and 2 is a local maximum.

Second Derivative Test

The Second Derivative Test serves as an alternative method to confirm the nature of critical points found through the First Derivative Test. Essentially, it uses the second derivative to determine the concavity of the function at those points.

Here's how it works:

• First, find the second derivative of the function.
• Evaluate it at each critical point:
• If the second derivative is positive, the function is concave up, indicating a local minimum.
• If it's negative, the function is concave down, indicating a local maximum.
• If the second derivative equals zero, the test is inconclusive.

In the exercise, evaluating the second derivative at the critical points confirmed the results of the First Derivative Test - 0 is a local minimum (concave up), and 2 is a local maximum (concave down).

Quotient Rule

The Quotient Rule is a fundamental tool for finding the derivative of a quotient of two functions. When you have a function in the form of \( \frac{u}{v} \), the derivative is given by\[ (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \].

Here's a simple breakdown:

• Identify the top function \( u \) and its derivative \( u' \).
• Identify the bottom function \( v \) and its derivative \( v' \).
• Substitute into the formula \( \frac{u'v - uv'}{v^2} \) to find the derivative.

For the original function \( f(x) = \frac{x^2}{x-1} \), using the quotient rule allows us to compute the first derivative, which is crucial for finding critical points and analyzing the function’s behavior. This info is used both in the First and Second Derivative Tests to confirm local extrema effectively.