Problem 79 - University Calculus: Early Transcendentals

Question

Find the absolute maximum value of \(f(x)=x^{2} \ln (1 / x)\) and say where it is assumed.

Step-by-Step Solution

Answer

The absolute maximum value is \(\frac{1}{2e}\) at \(x = e^{-1/2}\).

Step 1: Understand the domain of the function

The function is given as \(f(x) = x^2 \ln(1/x)\). Since \(\ln(1/x) = -\ln(x)\), we have \(f(x) = -x^2 \ln(x)\). The function is only defined for \(x>0\).

Step 2: Differentiate the function

We need to find the critical points to determine the absolute maximum. First, let's differentiate the function with respect to \(x\):\[f'(x) = \frac{d}{dx}(-x^2 \ln(x)) = -2x \ln(x) - x.\]

Step 3: Find the critical points

Set the first derivative to zero to find the critical points:\[-2x \ln(x) - x = 0.\]Factor out \(-x\):\[-x(2\ln(x) + 1) = 0.\]Since \(x eq 0\), the critical point is: \[2\ln(x) + 1 = 0 \Rightarrow \ln(x) = -\frac{1}{2} \Rightarrow x = e^{-1/2}.\]

Step 4: Evaluate function at critical point

Evaluate \(f(x)\) at the critical point \(x = e^{-1/2}\):\[f(e^{-1/2}) = (e^{-1/2})^2 \ln(\frac{1}{e^{-1/2}}) = e^{-1} \ln(e^{1/2}) = e^{-1} \times \frac{1}{2} = \frac{1}{2e}.\]

Step 5: Analyze behavior at boundaries

As \(x\to 0^+\), \(f(x)\to 0\) because \(x^2 \to 0\) faster than \(\ln(1/x) \to \infty\). As \(x \to \infty\), \(f(x)\to -\infty\) since \(\ln(1/x) \to -\infty\).

Step 6: Determine absolute maximum value

The only critical point found is \(x = e^{-1/2}\) and the boundaries show no maximum values. Therefore, the absolute maximum value of \(f(x)\) is \(\frac{1}{2e}\) at \(x = e^{-1/2}\).

Key Concepts

Derivative

In calculus, the derivative of a function helps us understand how the function's output changes as its input changes. For the function \( f(x) = x^2 \ln(1/x) \), we need to calculate the derivative to find the points where the function behaves in certain interesting ways, such as maximum or minimum points. The process involves differentiation, which is a way to systematically find the rate at which the function's value is changing at any given point.

• The derivative tells us the slope of the tangent to the curve at any point. If \( f'(x) > 0 \), the function is increasing at \( x \). If \( f'(x) < 0 \), it's decreasing.
• In our case, we differentiate \( f(x) = -x^2 \ln(x) \) to get \( f'(x) = -2x \ln(x) - x \).

This derivative helps us find critical points that are crucial for further analysis.

Critical points

Critical points are where the function's derivative is zero or undefined. These points are significant because they indicate potential locations for maximum, minimum, or saddle points.For the given function \( f(x) = -x^2 \ln(x) \), we set its derivative to zero to find the critical points. So, we solve:\[-2x \ln(x) - x = 0\]Factoring out \(-x\) from the equation yields:\[-x(2\ln(x) + 1) = 0\]Since \( x eq 0 \), the critical point is found by solving:\[ 2\ln(x) + 1 = 0 \Rightarrow \ln(x) = -\frac{1}{2} \Rightarrow x = e^{-1/2} \]This solution gives us a key point to further evaluate its behavior and importance in the function's graph.

Absolute maximum

The absolute maximum of a function is the highest point over its entire domain, showing where the output value is the largest.After finding the critical points, we evaluate the original function \( f(x) \) at these points and also consider the behavior at the domain's borders. For our function, we found a critical point at \( x = e^{-1/2} \).

• We calculated \( f(e^{-1/2}) = \frac{1}{2e} \).

To ensure this is the absolute maximum:

• We check where \( x \to 0^+ \), the function approaches 0 because \( x^2 \to 0 \) faster than \( \ln(x) \to -\infty \).
• As \( x \to \infty \), \( f(x) \to -\infty \), showing no competition for a higher value than at \( x = e^{-1/2} \).

Thus, \( \frac{1}{2e} \) is the absolute maximum value of \( f(x) \), achieved at \( x = e^{-1/2} \).

Function analysis

Analyzing a function involves understanding its behavior, trends, and peculiar points over its domain. We look at aspects such as continuity, differentiability, critical points, and boundary behaviors.For \( f(x) = -x^2 \ln(x) \):

• The function is defined only for \( x > 0 \).
• We found that as \( x \to 0^+ \), the function nears 0, implying no absolute minimum in this direction.
• The critical point \( x = e^{-1/2} \) has been established as a potential maximum.
• Behavior at \( x \to \infty \) shows the function dropping to \(-\infty\).

These insights give us a full picture, allowing us to conclude about the function's characteristics and the location of its absolute maximum.