The function logx-x is increasing in the interval

The correct option is A (0, e) f'(x)=1x^2 log xx^2 0=(1 log) xx^2 x=e f'(x) gt;0 #160; when x∈ (0, e) ...

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Logarithmic Inequalities

The function ...

Question

The function $\frac{log x}{x}$ is increasing in the interval

(1, 2 e)

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(0, e)

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(e, 2 e)

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(\frac{1}{e}, 2 e)

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