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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 7.2.71b

71. Locate and identify the absolute extreme values of cos(ln x) on [1/2, 2]

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First, understand that the function to analyze is \(f(x) = \cos(\ln x)\) on the interval \(\left[ \frac{1}{2}, 2 \right]\). We want to find the absolute maximum and minimum values on this closed interval.
Step 1: Find the derivative of the function to locate critical points. Using the chain rule, the derivative is \(f'(x) = -\sin(\ln x) \cdot \frac{1}{x}\).
Step 2: Set the derivative equal to zero to find critical points inside the interval: \(-\sin(\ln x) \cdot \frac{1}{x} = 0\). Since \(\frac{1}{x} \neq 0\) for \(x > 0\), this reduces to \(\sin(\ln x) = 0\).
Step 3: Solve \(\sin(\ln x) = 0\) for \(x\) in \(\left[ \frac{1}{2}, 2 \right]\). Recall that \(\sin t = 0\) when \(t = k\pi\) for any integer \(k\). So, set \(\ln x = k\pi\) and solve for \(x = e^{k\pi}\), then determine which values lie in the interval.
Step 4: Evaluate the function \(f(x) = \cos(\ln x)\) at all critical points found in the interval and also at the endpoints \(x = \frac{1}{2}\) and \(x = 2\). Compare these values to identify the absolute maximum and minimum.

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Critical points occur where the derivative of a function is zero or undefined. Finding these points helps identify potential locations of extreme values within the interval.
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