Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • (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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App