Find how many solutions does $f(x)=\ln x-kx$ has for $k>\frac 1 e$.
$f<0$ at $x\to \infty$ and $x\to 0$. The derivative has a solution only at $x=\frac 1 k$.
So place that point in $f$ and we'll want to check when $f(\frac 1 k)>0$ and $f(\frac 1 k)<0$
Now I'm getting confused here:
$f(\frac 1 k)>0\to \ln \frac 1 k > 1$ so $e^1>\frac 1 k\to k>\frac 1 e$ but it should be the opposite, so my question is why does the inequality 'flip'?