Problem 26 - Calculus Single Variable

Question

Use the first derivative to determine the intervals on which the given function \(f\) is increasing and on which \(f\) is decreasing. At each point \(c\) with \(f^{\prime}(c)=0,\) use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=x^{2} \ln (x) $$

Step-by-Step Solution

Answer

Function is decreasing on \((0, \frac{1}{\sqrt{e}})\) and increasing on \((\frac{1}{\sqrt{e}}, \infty)\); \(f\) has a local minimum at \(x = \frac{1}{\sqrt{e}}\).

Step 1: Find the First Derivative

To find where the function is increasing or decreasing, we first need to calculate the first derivative of the function \( f(x) = x^2 \ln(x) \). We use the product rule here, where if \( u = x^2 \) and \( v = \ln(x) \), then \( f'(x) = u'v + uv' \). Therefore, \( f'(x) = 2x \ln(x) + x^2 \cdot \frac{1}{x} = 2x \ln(x) + x \).

Step 2: Solve for Critical Points

To find critical points, set \( f'(x) = 0 \). That is, solve \( 2x \ln(x) + x = 0 \). Factor out \( x \): \( x (2 \ln(x) + 1) = 0 \). Thus, \( x = 0 \) or \( 2 \ln(x) + 1 = 0 \). Since \( x = 0 \) is not in the domain of \( \ln(x) \), solve \( 2 \ln(x) + 1 = 0 \): \( \ln(x) = -\frac{1}{2} \), so \( x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \).

Step 3: Determine Intervals

We need to analyze the sign of \( f'(x) = 2x \ln(x) + x \) on the intervals formed by the critical point \( x = \frac{1}{\sqrt{e}} \). The critical point divides the domain into \( (0, \frac{1}{\sqrt{e}}) \) and \( (\frac{1}{\sqrt{e}}, \infty) \).

Step 4: Test Intervals for Monotonicity

Choose a test point from each interval. For \( (0, \frac{1}{\sqrt{e}}) \), choose \( x = \frac{1}{4} \): \( f'(\frac{1}{4}) = 2(\frac{1}{4}) \ln(\frac{1}{4}) + \frac{1}{4} < 0 \), indicating \( f(x) \) is decreasing on this interval. For \( (\frac{1}{\sqrt{e}}, \infty) \), choose \( x = 1 \): \( f'(1) = 2(1) \ln(1) + 1 = 1 > 0 \), indicating \( f(x) \) is increasing on this interval.

Step 5: Classify the Critical Point Using the First Derivative Test

At \( x = \frac{1}{\sqrt{e}} \), we have a sign change in \( f'(x) \) from negative to positive, suggesting a local minimum. Thus, \( f(\frac{1}{\sqrt{e}}) \) is a local minimum value.

Key Concepts

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. They play a crucial role in determining where a function might have a local maximum or minimum. In our problem, we have the function \( f(x) = x^2 \ln(x) \). To find critical points, we first compute the first derivative \( f'(x) \) using the product rule and set it to zero: \( f'(x) = 2x \ln(x) + x = 0 \). This gives us potential critical points by considering two cases: \( x = 0 \) or \( 2 \ln(x) + 1 = 0 \).Because \( x = 0 \) does not exist in the domain of the logarithm, we solve \( 2 \ln(x) + 1 = 0 \) leading to \( x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \). This is our critical point, and it's within the function's domain.

Monotonicity

Monotonicity refers to where a function is increasing or decreasing. We use the first derivative to determine this feature by evaluating the sign of \( f'(x) \). For \( f(x) = x^2 \ln(x) \), the first derivative is \( f'(x) = 2x \ln(x) + x \). We test this derivative at intervals around the critical point.

• For the interval \((0, \frac{1}{\sqrt{e}})\), we select a test point like \( x = \frac{1}{4} \). The derivative \( f' \left( \frac{1}{4} \right) = 2 \times \frac{1}{4} \times \ln \left( \frac{1}{4} \right) + \frac{1}{4} \) results in a negative value, indicating decreasing behavior here.
• For the interval \((\frac{1}{\sqrt{e}}, \infty)\), using a test point such as \( x = 1 \), gives \( f'(1) = 2 \times 1 \times \ln(1) + 1 = 1 \). This result is positive, showing the function is increasing.

Thus, the function transitions from decreasing to increasing across the critical point.

Product Rule

The product rule is essential when differentiating products of functions. For a function \( f(x) = u(x) \cdot v(x) \), the derivative is \( f'(x) = u'(x)v(x) + u(x)v'(x) \). In our case, \( u(x) = x^2 \) and \( v(x) = \ln(x) \).

• \( u'(x) = 2x \) since the derivative of \( x^2 \) is \( 2x \).
• \( v'(x) = \frac{1}{x} \) as the derivative of \( \ln(x) \) is \( \frac{1}{x} \).

Applying the product rule therefore gives \( f'(x) = 2x \ln(x) + x^2 \cdot \frac{1}{x} = 2x \ln(x) + x \). Understanding the product rule helps us correctly find \( f'(x) \) which is crucial for determining critical points and monotonicity.

Local Minimum

A local minimum occurs at a point where a function value is lower than all nearby points. The First Derivative Test is a tool to identify such points by inspecting the sign change of the derivative.At the critical point \( x = \frac{1}{\sqrt{e}} \), the derivative changes from negative on \( (0, \frac{1}{\sqrt{e}}) \) to positive on \( (\frac{1}{\sqrt{e}}, \infty) \). This transition from decreasing to increasing indicates a local minimum.Therefore, at \( x = \frac{1}{\sqrt{e}} \), the function \( f(x) = x^2 \ln(x) \) reaches a local minimum value. Understanding this concept is key for many practical applications, such as optimization problems. It's a powerful way to find pockets of low points within a function's interval.