I've got the following function: $$f(x)=\frac{1}{16x}-\frac{1}{(x+3)^2} $$ And I wish to show that it is convex in the open interval $(0,3)$, took the second derivative, i.e.$$f''(x)=\frac{1}{8x^3}-\frac{6}{(x+3)^4}$$ I painstakingly tried to expand it and algebraically manipulate it to solve $f''(x)>0$, but that seems pretty hopeless, how should I go about this?


We are claiming that

$$f''(x)=\frac{1}{8x^3}-\frac{6}{(x+3)^4}>0 \iff\frac{(x+3)^4-48x^3}{8x^3(x+3)^4}>0$$

then we need to prove that (using the idea by Timothy for the rational roots)

$$\bar g(x)=(x+3)^4-48x^3=(x-3)(x^3-33x^2-45x-27)>0$$

that is


and we have

  • $g(0)=-27$
  • $g(3)=-432$
  • $h(x)=g'(x)=3x^2-66x-45<0$


  • $g(x)$ is strictly decreasing on the interval
  • $g(x)$ is negative on the interval

Here's something I would do:

Solve $f''(x)=0$ and test a value inside your open interval.

So we have: $$\frac1{8x^3}-\frac6{(x+3)^4}=0$$


Then I would cross multiply and expand $(x+3)^4$:


$$x^4+12x^3+54x^2+108x+81=48x^3$$ $$x^4-36x^3+54x^2+108x+81=0$$

At this point, we could use the Rational Root Theorem to guess roots.

The possible roots are $\pm1, \pm3, \pm27, \pm81$.

Of these, we find that $x=3$ works, as $3^4-36(3)^3+54(3)^2+108(3)+81=0$.

I think you could take it from here.

  • $\begingroup$ Damn! I tried to apply RRT, checked 1, and -1 and was like, yeah, this won't work, turns out, I shouldn't have been lazy -.- $\endgroup$ – Spasoje Durovic Nov 24 '18 at 18:33
  • $\begingroup$ yeah sometimes the roots are like 3 or -3, I was kind of able to tell that it was 3 because the coefficients had a common factor of 3 $\endgroup$ – Timothy Cho Nov 24 '18 at 18:33
  • 1
    $\begingroup$ RRT only gives us the rational roots but how can we conclude form here that we have not others not rational roots on the interval $(0,3)$? $\endgroup$ – gimusi Nov 24 '18 at 18:40
  • $\begingroup$ I'm trying to complete the arguement, anyway, I've read through your answer and it's quite nice, I don't think I would've came up with that, so I'm trying to adapt my initial attempt, if i'm unable to, I will try to go carefully through yours to get more insight on how you came up with it $\endgroup$ – Spasoje Durovic Nov 24 '18 at 18:46
  • 1
    $\begingroup$ @SpasojeDurovic The edited version is simpler than the forst one. For the previous the reasoning make use of simple properties for continuos funtions (IVT and Weierstrass) and for derivatives. We have that the previsous g(x) was equal to 0 ate the limits of the interval. Now, since the derivative h(x) has exactly one root on that interval, considering also the sign of h(x), g(x) must have exactly a maximum and then it can cross the x axis, otherwise g(x) would have also a minimum on that interval. $\endgroup$ – gimusi Nov 24 '18 at 19:16

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.