Show that the function given by f x -log x - x has maximum at x - e-

Let f(x) = log x/x On differentiating w.r.t.x, we get f'(x) = x(1/x) (log x).1/x^2 = 1 log x/x^2 Again differentiating, we get f"(x) = x^2( 1/x) (1 log x)2x ...

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Standard XII

Mathematics

Definite Integral as Limit of Sum

Show that the...

Question

Show that the function given by $f (x) = \frac{l o g x}{x}$ has maximum at x = e.

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Solution

Let $f (x) = \frac{l o g x}{x}$
On differentiating w.r.t.x, we get $f^{'} (x) = \frac{x (\frac{1}{x}) - (l o g x) .1}{x^{2}} = \frac{1 - l o g x}{x^{2}}$
Again differentiating, we get $f " (x) = \frac{x^{2} (- \frac{1}{x}) - (1 - l o g x) 2 x}{(x^{2})^{2}}$
$= \frac{- x - 2 x + 2 x l o g x}{x^{4}} = \frac{x (2 l o g x - 3)}{x^{4}} = \frac{2 l o g x - 3}{x^{3}}$
For maximum put $f^{'} (x) = 0 \Rightarrow \frac{1 - l o g x}{x^{2}} = 0 \Rightarrow l o g x = 1 \Rightarrow x = e$
At $x = e, f " (e) \frac{2 l o g e - 3}{e^{3}} = \frac{2.1 - 3}{e^{3}} = \frac{- 1}{e^{3}} < 0$
Therefore, by second derivative test, f is the maximum at x = e.

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Show that the function given by f(x)=log xx has maximum at x = e.

Show that the function given by $f (x) = \frac{l o g x}{x}$ has maximum at x = e.