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Is the following statement true?

Suppose we are given $f(x)$ defined and differentiable $\forall x \in \mathbb{R}$, $x\neq 0$. If $xf'(x)>0$, then $f$ has no minimum value.

The statement seems reasonable, because if I choose $f(x) = -1/x^2$, then its derivative is $1/x^3$ so $xf'(x) = 1/x^2$ that is $>0$, for all $x\ne0$.

And $-1/x^2$ clearly has no minimum value (its lower bound is $-\infty$).

My problem is that I have no idea how I should continue going from this example to a general proof.


Taken from a question of a calculus 1 exemption of 2011.

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  • $\begingroup$ It is false. Take f(x) =|x| $\endgroup$ Dec 26, 2017 at 9:49
  • $\begingroup$ @GiuseppeNegro |x| is defined in x=0 that it cannot be $\endgroup$ Dec 26, 2017 at 10:47

2 Answers 2

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$xf'(x) > 0 $ tells us that $f$ is decreasing when $x<0$ and $f$ is increasing when $x>0$. This must mean that if there is a minimum, it must be at $x=0$. But $x=0$ is not included in the domain, so that $f$ has no minimum.

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  • $\begingroup$ If the derivative of f(x) is f'(x) = x, then we know that f is decreasing for x < 0 and increasing for x > 0, but if the derivative f'(x) is different from 1, how can you say that from xf'(x) > 0 that f is decreasing for x < 0 and increasing for x > 0? $\endgroup$ Dec 24, 2017 at 12:34
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    $\begingroup$ If $x>0$ and $xf'(x)>0$, it must mean that $f'(x)>0$, so $f$ is increasing. And for $x<0$ and $xf'(x)>0$, it must mean that $f'(x)<0$, so $f$ is decreasing. $\endgroup$
    – Higurashi
    Dec 24, 2017 at 12:37
  • $\begingroup$ in the second did you mean $xf'(x) > 0$? because what you wrote seems to go against the problem statement $\endgroup$ Dec 24, 2017 at 12:41
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    $\begingroup$ Yes, I've edited that, sorry :). $\endgroup$
    – Higurashi
    Dec 24, 2017 at 12:42
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If $f$ has a minumum at $x_0$ then $f'(x_0)=0$. Hence, $0<x_0f'(x_0)=0$, a contradiction.

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