Question

find the minimum value of x/log x

Answer 1

Answer:

Minimum value = e

Step-by-step explanation:

At local minima first derivative of function is zero and second derivative is positive.

$f(x) = \frac{x}{log(x)} \\ \frac{dy}{dx} = \frac{log(x) \times 1 - \frac{1}{x} \times x }{{( \ \log(x))}^{2} } \\ = \frac{log(x) - 1}{ {(log(x))}^{2} } \\ and \: second \: derivtive \: is \\ \frac{ {d}^{2} y}{d {x}^{2} } = \frac{ {(log(x))}^{2} \times \frac{1}{x} - (log(x) - 1) \times \frac{2log(x)}{x} }{ {(log(x))}^{4} }$

Now equate first derivative to zero

And at that point verify if second derivative is positive

$\frac{log(x) - 1}{ {(log(x))}^{2} } = 0 \\ logx = 1 \\ i.e. \: x = e \\ at \: x = e \: second \: derivative \: is \\ = \frac{1}{e} > 0$

Therefore f (x) is minimum at x=e

Now minimum value of the function is f (e) = e/log (e) = e

Hope this helps you

Answer 2

Answer:

the local minimum value of x/logx = e

Step-by-step explanation:

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