Find the intervals in which the function \( f(x) = \frac{\log x}{x} \) is strictly increasing or strictly decreasing.
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Solution and Explanation
1. Find the derivative: \[ f(x) = \frac{\log x}{x}, \quad f'(x) = \frac{1 \cdot x - \log x \cdot 1}{x^2} = \frac{x - \log x}{x^2}. \]
2. Critical points: The critical points are obtained by solving \( f'(x) = 0 \): \[ x - \log x = 0 \quad \Rightarrow \quad x = \log x. \] Let \( x = e^k \). The equality holds for \( x = e \).
3. Sign of \( f'(x) \): - For \( x \in (0, e) \), \( x - \log x>0 \), so \( f'(x)>0 \): \( f(x) \) is increasing. - For \( x \in (e, \infty) \), \( x - \log x<0 \), so \( f'(x)<0 \): \( f(x) \) is decreasing.
Intervals: \[ f(x) { is strictly increasing on } (0, e) { and strictly decreasing on } (e, \infty). \]
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R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
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