# High school math question $f(x)=ax+2$, $g(x)=a^2x^2-\ln x+2$ $a\in \mathbb{R}$, $x>0$.

Q: Is there a negative $a$, for any positive $x$, $f(x)\le g(x)$? If $a$ exist, solve it, else, show the reason.

I want to know how to solve this problem, and are there any soft wares to show the figure of $ax-a^2x^2+\ln x\le 0$?

• Write it as $f(x) - g(x)$ and compare to 0. – Bartek Banachewicz Jul 29 '13 at 8:58

It is easy enough to see that you can find an $a$ and an $x$ to satisfy this by looking at what happens around $x = 0$. As $x$ approaches $0$, $f(x)$ goes towards $2$ and $g(x)$ goes towards infinity (due to the $\ln(x)$ component). That makes it easy to find a negative a that for small $x$ makes the $f(x) \le g(x)$.