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I would appreciate if somebody could help me with the following problem

Q: prove that ($f:\mathbb{R}\to \mathbb{R}$)

if $f(x)+xf'(x)>0$ then $f(x)>0$

Young
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    $$f(x)+xf'(x)=\frac{d\left(xf(x)\right)}{dx}$$ – lab bhattacharjee Dec 22 '13 at 16:41
  • more hint Please! – Young Dec 22 '13 at 16:44
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    Are you assuming this holds on an interval or on all of $\Bbb R$? It would help you in thinking about the problem (and in using the hints) to know exactly what your hypotheses are and whether you're using them. – Ted Shifrin Dec 22 '13 at 16:44
  • The intuition behind this problem is clearly that if we have a minimum, then by Fermat's rule, if $x_0$ is the minimizer, then $f(x_0) = f(x_0) + x_0 f'(x_0) > 0$, so $f$ must be strictly positive everywhere. For some reason I can't find any proof that $f(x) + x f'(x) > 0$ implies the existence of a minimum... any ideas people? I believe this is interesting. – Patrick Da Silva Dec 22 '13 at 17:01

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hint this $$(xf(x))'=f(x)+xf'(x)$$

so let $$F(x)=xf(x)$$ then $$F'(x)>0$$ so if $x\ge0$,then $$F(x)>F(0) =0\Longrightarrow xf(x)>0\Longrightarrow f(x)>0$$ if $x\le 0$,then $$F(x)\le 0\Longrightarrow f(x)>0$$

math110
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