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 $ax-a^2x^2+\ln x\le 0$?

  • $\begingroup$ Write it as $f(x) - g(x)$ and compare to 0. $\endgroup$ – 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 $f(x) \le g(x)$.

  • $\begingroup$ Your method is good ,would you please give the precise solution? and what is the arange of a ? $\endgroup$ – Aostio Jul 29 '13 at 9:01
  • $\begingroup$ Not sure I understand quite what you're after. If you want to find all the possible values of a for any given x then you can just solve the quadratic in your last equation. I'm not competent enough with the markup of this site to be able to write that in any readable way though. $\endgroup$ – Chris Jul 29 '13 at 9:50
  • $\begingroup$ Would you please show us a picture or screenshot about the procedures how you solve the question -- Is there a negative a, for any positive x, f(x)≤g(x)always true? If a exist, solve it, else, show the reason . $\endgroup$ – Aostio Jul 29 '13 at 10:02

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

  • $\begingroup$ Thanks for you suggestion.I will try. $\endgroup$ – Aostio Jul 29 '13 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.