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 $ax-a^2x^2+\ln x\le 0$?
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 $f(x) \le g(x)$.
Sorry I can't comment yet, but why don't you just try graphing it in mathematica or even on WolframAlpha if you don't have access to mathematica. Especially useful is mathematica's variable plot function which allows you to graph functions and then watch the graphs change as you vary a parameter, such as a or x. You could even make two variables that can be manually changed and check if f(x) is ever less than g(x).

Mathematica is often your best bet for graphing functions and stuff