Showing that $f(x)$ is convex in $(0,3)$

Question

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?

Answer 1

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

$$g(x)=x^3-33x^2-45x-27<0$$

and we have

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

therefore

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

Answer 2

Here's something I would do:

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

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

$$\frac{1}{8x^3}=\frac{6}{(x+3)^4}$$

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

$$(x+3)^4=48x^3$$

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

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.