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.
$\endgroup$ 22 Answers
$\begingroup$$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.
$\endgroup$ 4 $\begingroup$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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