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Prove that $\log _e x $ is strictly increasing function on $(0 ,\infty)$.

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  • (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
  • (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$
Let $f(x)=\log _ex$
In the open , interval $ (0, \infty) \; f(x)$ is differentiable
$ f'(x) =\large\frac{1}{x} $ in $(0, \infty)$
and $\large\frac{1}{x} >$$0$ for $ x \in (0, \infty)$
$\therefore \log _ex$ is strictly increasing on $(0, \infty)$
answered Jul 29, 2013 by meena.p
 

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