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

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Rewrite the function to a more convenient form: \(f(x) = x^2 \ln\left(\frac{1}{x}\right) = x^2 (-\ln x) = -x^2 \ln x\) for \(x > 0\) since \(\ln\left(\frac{1}{x}\right)\) is defined only for positive \(x\).

Find the first derivative \(f'(x)\) using the product rule. Let \(u = -x^2\) and \(v = \ln x\), then \(f(x) = u v\). Compute \(f'(x) = u' v + u v'\) where \(u' = -2x\) and \(v' = \frac{1}{x}\).

Set the derivative equal to zero to find critical points: \(f'(x) = 0\). Solve the resulting equation for \(x\) to find candidates for local maxima or minima.

Evaluate the function \(f(x)\) at the critical points found and also consider the behavior of \(f(x)\) as \(x\) approaches the boundaries of the domain (i.e., as \(x \to 0^+\) and as \(x \to \infty\)) to determine the absolute maximum.

Compare the values of \(f(x)\) at the critical points and the limits to identify the absolute maximum value and the point where it is assumed.

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주요 개념

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Domain and Behavior of the Function

Understanding the domain of f(x) = x^2 * ln(1/x) is crucial since ln(1/x) is defined only for x > 0. Analyzing the behavior near the boundaries (as x approaches 0 and infinity) helps identify potential maximum values and ensures the function is well-defined where we search for extrema.

Finding Critical Points Using Derivatives

To find absolute maxima, compute the derivative f'(x) and solve f'(x) = 0 to locate critical points. These points are candidates for local maxima or minima. Differentiation involves applying the product and chain rules to handle the x^2 and ln(1/x) terms.

Evaluating Absolute Maximum Values

After finding critical points, evaluate f(x) at these points and at the domain boundaries to determine the absolute maximum. Comparing these values identifies where the function attains its highest value, ensuring a complete analysis of maxima.

133. Find the absolute maximum value off(x) = x^2 * ln(1/x)and say where it is assumed.