Solveeit Logo
Login

Question

Mathematics Question on Maxima and Minima

The maximum value of f(x)=log⁡xx,0<x<∞ f(x) = \frac{\log x}{x} , 0 < x < \infty is

A

2e\frac{2}{e}

B

1e\frac{1}{e}

C

e\sqrt{e}

D

ee

Answer

1e\frac{1}{e}

Explanation

Solution

We have, f(x)=log⁡xx,0<x<∞ f(x) = \frac{\log x}{x} , 0 < x < \infty
Maximum or minimum point is given by f′(x)=0f'(x) =0
f′(x)=x1x−log⁡x.1x2=0⇒1−log⁡xx2=0f'\left(x\right) =\frac{x \frac{1}{x} -\log x.1}{x^{2}} = 0 \Rightarrow \frac{1-\log x}{x^{2}} =0
⇒1=log⁡x⇒x−e\Rightarrow 1=\log x \Rightarrow x-e
Now f"(x)=x2(−1x)−(1−log⁡x)2xx4f" \left(x\right) =\frac{x^{2} \left(\frac{-1}{x}\right) -\left(1-\log x\right)2x}{x^{4}}
                =−1−2+2log⁡xx3=−3+2log⁡xx3f"(x)x=e=−1e3<0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{-1-2+2\log x}{x^{3}} =\frac{-3+2\log x}{x^{3}} f"\left(x\right)_{x=e} =\frac{-1}{e^{3} } <0
⇒x=e\Rightarrow x = e is a maximum.point and maximum value of f(x)f(x) is given by , f(e)=log⁡ ee=1ef(e) = \frac{\log \: e}{e} = \frac{1}{e}