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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.PE.116

In Exercises 115 and 116, find the absolute maximum and minimum values of each function on the given interval.
116. y = 10x (2 - ln(x)), (0, e²]"133. Find the absolute maximum value of
f(x) = x^2 * ln(1/x)
and say where it is assumed

Guida verificata passo dopo passo
1
First, rewrite the function to a more convenient form. Given \( f(x) = x^2 \ln\left(\frac{1}{x}\right) \), use the logarithm property \( \ln\left(\frac{1}{x}\right) = -\ln(x) \) to rewrite it as \( f(x) = x^2 (-\ln(x)) = -x^2 \ln(x) \).
Next, find the critical points by computing the derivative \( f'(x) \). Use the product rule on \( -x^2 \ln(x) \): \( f'(x) = -\left( 2x \ln(x) + x^2 \cdot \frac{1}{x} \right) = -\left( 2x \ln(x) + x \right) \).
Set the derivative equal to zero to find critical points: \( f'(x) = 0 \Rightarrow -\left( 2x \ln(x) + x \right) = 0 \Rightarrow 2x \ln(x) + x = 0 \). Factor out \( x \): \( x(2 \ln(x) + 1) = 0 \). Since \( x > 0 \), solve \( 2 \ln(x) + 1 = 0 \) for \( x \).
Solve for \( x \) from the equation \( 2 \ln(x) + 1 = 0 \): \( \ln(x) = -\frac{1}{2} \), so \( x = e^{-1/2} \). This is the critical point inside the domain.
Evaluate \( f(x) \) at the critical point \( x = e^{-1/2} \) and at the endpoints of the domain (if given) to determine the absolute maximum value and where it is assumed.

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