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One of my friends brother asked me this one, Let $f$ be a real valued function of a variable $x$ such that $f'(x)$ takes both positive and negative values and $f''(x)>0$ for all $x$. To show there is a real number $p$ such that $f$ is increasing $\forall x\ge p$. What I tried to make him understand that the function is convex and geometrically I got the idea but failed to give him a clear solution.

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    If f''>0 for all $x$, then $f'$ is increasing for all $x$. Since $f'$ takes a positive value, say at $x=a$, f'(x)>0 for all $x\ge a$; and thus, $f$ is increasing on $(a,\infty)$.2012-04-16

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Assuming the domain of $f$ is $\Bbb R$:

If $f''(x)>0$ for all $x\in\Bbb R$, then $f'$ is increasing for all $x\in\Bbb R$. Since $f'$ takes a positive value, say at $x=a$, $f'(x)>0$ for all $x\ge a$; and thus, $f$ is increasing on $(a,\infty)$.

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    I was just about to suggest that you make that an answer. Maybe even add that it’s enough to have some $x_0$ such that $\operatorname{dom}f\supseteq (x_0\to)$ and $f\,'$ takes on both positive and negative values on $(x_0,\to)$.2012-04-16
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How about $f(x) = -\log x$, so that $f'(x) = -1/x$ and $f''(x)=1/x^2$. Then $f''>0$ on $(0,\infty)$ but $f$ is decreasing on that whole interval.

Later edit: It's being pointed out that the "OP" said that $f'$ should be positive at some point.

Suppose $f'(c)>0$. Then, since $f''>0$ everywhere, we conclude that $f'$ is increading everywhere. If it's increasing everywhere, and positive at $c$, then it's positive everywhere in $[c,\infty)$. And so $f$ is increasing everywhere in $[c,\infty)$. So let the $p$ that the "OP" asks for be this number $c$.

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    If we change $f(x)$ to be $-\log|x|$, then the hypothesis is satisfied (but the function is no longer continuous).2012-04-16
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Let $f'(x)=-2-x^{-1}$. Then, indeed, $f''(x)>0$, but there is no $x>0$ such that $f'(x)>0$.

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    You had $x^-1$ where you presumably needed $x^{-1}$. The $\TeX$ code for the former is x^-1 and for the latter is x^{-1}. Generally if there's just one character in the superscript, the curly braces would be superfluous, e.g.$x^2$gives you $x^2$. With more than one character in the superscript, the braces are needed.2012-04-16