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Prove that $e^{x} $ is strictly increasing function on $R$.

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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 $f(x)=e^x$ Which is defined and differentiable on $R$
$f'(x) =e^x >0$ for all $ x\in R$
$\therefore e^x$ is strictly increasing on $R$


answered Jul 29, 2013 by meena.p

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