# Minimum of $\frac{x^3}{x-6}$ for $x>6$ without using derivative?

Find the minimum of $$y=f(x)=\dfrac{x^3}{x-6}$$ for $$x>6$$.

I can solve the question using derivatives but I have no any idea how to do it without them. Using derivatives, we find $$x=9$$ and $$y_{min}=243$$.

When looking for the minimum of a function, calculus is the default choice (with good reason), but it is a relatively new idea in Mathematics. Questions such as the one posed here are an interesting intellectual challenge because the obvious approach (calculus) is not the only approach. Without calculus there is no obvious approach though...

• What's the problem of solving it with the derivative? is it not allowed? Commented Apr 19 at 12:41
• Yes, it is not allowed. Commented Apr 19 at 12:49
• Possible hint: long division. Commented Apr 19 at 12:49
• Why is using the derivative not allowed? I ask because the rationale behind it also might accidentally exclude other approaches. Or the rationale might point to a reason to it one way or another. Commented Apr 19 at 21:05
• Notice that $\frac{x^3}{x-6}=243+\frac{(x-9)^2(x+18)}{x-6}\ge243$. Commented Apr 21 at 11:37

Here is an alternative approach.

Assume that $$y=\frac{x^{3}}{x-6}$$ has a minimum value of $$y=c$$ where $$c\in R$$.

Then $$\frac{x^3}{x-6}-c=0$$ will have a repeated root somewhere. Define this point as $$x=a$$.

This means $$x^3-c(x-6)=0$$ can be written in the form $$(x-b)(x-a)^2$$ and we are told that $$a>6$$.

Expanding and comparing coefficients gives $$2a+b=0, a^2+2ab=-c$$ and $$-a^2b=6c$$

Solving these for $$a$$ and $$c$$ gives either $$a=0$$ (which we reject since $$a>6$$) and $$a=9$$ as well as $$c=3a^2$$ which means $$c=243$$.

So the graph has a minimum at $$(9,243)$$.

• Does $y=c$ line intersect the function at $x=a$ point Red Five? Commented Apr 30 at 14:11
• Should be a tangent at the point given the shape of the graph. Commented Apr 30 at 20:19

Applying the AM-GM inequality, we have: \begin{align*} \dfrac{x^3}{x-6} &= 108+\dfrac{216}{x-6}+18(x-6)+(x-6)^2 \\ &= \begin{aligned}[t]&108+ \dfrac{27}{x-6}+\dfrac{27}{x-6}+\dfrac{27}{x-6}+\dfrac{27}{x-6}+\dfrac{27}{x-6}+\dfrac{27}{x-6}+\dfrac{27}{x-6}+\dfrac{27}{x-6} \\ &+3(x-6)+3(x-6)+3(x-6)+3(x-6)+3(x-6)+3(x-6)+(x-6)^2\end{aligned}\\ &\ge 108 + 15\sqrt[15]{\left(\dfrac{27}{x-6}\right)^8(3(x-6))^6(x-6)^2} \\ &=108+15\sqrt[15]{27^8\cdot 3^6}\\ &=108+15\sqrt[15]{3^{24}\cdot 3^6}\\ &=108+15\sqrt[15]{3^{30}}=108+15\cdot 3^2\\ &=108+135=243 \end{align*} Equality occurs when $$\dfrac{27}{x-6} = 3(x-6) = (x-6)^2\implies x-6 = 3\implies x = 9.$$ Thus $$y_{\mathrm{min}} = 243$$.

One thing you can do is recognize that at the minimum value, $$x$$, the function must be locally convex, so that a small perturbation $$t$$ to the left and right of $$x$$ should result in the same value as $$t \rightarrow 0$$.

Therefore, we can say that as $$t \rightarrow 0$$, it must be that $$\frac{(x+t)^3}{x+t-6} + \frac{(x-t)^3}{x-t-6} = 2\frac{(x+t)^3}{x+t-6}.$$

If we solve this equation for $$t$$ in terms of $$x$$, we find that

$$t = \pm x\sqrt{\frac{x-9}{x+3}}.$$

So, for this equation to hold as $$t \rightarrow 0$$, we can see that $$x$$ must be equal to 9.

• Can you say a bit more on how to get the value of $t$ from the first equation? I'm missing something =) Commented Apr 21 at 19:18
• I just put the first equation into wolfram alpha and asked to solve for t. It might involve solving a quartic equation, which is difficult but not impossible.
– Doug
Commented Apr 22 at 13:22
• I see, thanks =) Commented Apr 23 at 9:02
• This is just differentiation really. Commented Apr 25 at 11:49

Here is a direct proof that $$\forall x>6\quad f(x)\ge f(9)$$, which can be first guessed by a plot: \begin{align}\forall h>-3,&\quad\frac{(9+h)^3}{3+h}\ge\frac{9^3}3\\ &\iff(9+h)^3\ge3^5(3+h)\\ &\iff h^2(h+27)\ge0. \end{align}

Afterthought: another way than a plot (to guess $$9$$ is the place of minimum before proving it like above) is the following. If $$f$$ has a local minimum at some $$x>6$$ then, for every $$t$$ such that $$|t|$$ is small enough, $$(x+t)^3(x-6)\ge x^3(x+t-6),$$ i.e. $$t(2x^3-18x^2)+t^2(3x^2-18x)+t^3(x-6)\ge0.$$ For this to hold for $$|t|$$ small enough whatever the sign of $$t$$, $$2x^3-18x^2$$ must be $$0$$.

• Anne, you participate in closing so many problem statement "no clue" questions, and yet you choose to answer this one? Commented Apr 22 at 1:29
• @amWhy Not sure indeed whether this challenge "deserved" answers, but I am not a robot. I needed some (harmless) fun. I think all these comments (including the present one) should now be deleted. Commented Apr 22 at 6:19

Using Am-Gm inequality for 3 (nonnegative) summands $$a+b+c \geq 3\cdot \sqrt[3]{abc}$$

If $$t=x-6>0$$ then \begin{align}f(x) &= {(t+6)^3\over t}\\ &= {(t+3+3)^3\over t}\\ &\geq {(3\sqrt[3]{9t})^3\over t}\\ &= 243\\ \end{align}

with equality iff $$t=3$$ i.e. $$x= 9$$.

As an alternative to Red Five nice solution, assuming that we have a minimum $$m$$ (which is true by EVT):

$$\frac{x^3}{x-6}=m \iff x^3-mx+6m=0$$

which is a depressed cubic equation in the form $$x^3+px+q=0$$ with multiple roots, which leads to

$$\Delta =4p^3+27q^2=0 \implies 4m^3-972m^2=0 \implies m=243$$

• Very nice. This would mean that it is possible to make a more general rule, provided the fraction is $\frac{x^{3}}{ax+b}$ would it not? Commented May 2 at 1:51
• @RedFive Yes if we are sure that a minimum exists as in this case.
– user
Commented May 2 at 5:36

In the whole domain $$\inf{f(x)}=-\infty$$. Let's check for $$x>6$$ since we have an asymptote there, by calculating a few values by hand we find out that: $$f(7)=343, f(8)=256,f(9)=243,f(10)=250$$. Since we know that $$f(x)$$ is increasing as $$x$$ goes to $$\infty$$ the minimum must be inside $$[8,10]$$ Let's split this interval this way: $$I_1=[8,\frac{8+10}{2}]\cup J_1=[\frac{8+10}{2},10]\Rightarrow I_1\cup J_1=[8,9]\cup[9,10]$$ Let's evaluate the function at these $$3$$ values, and then we'll split the intervals again in the same way where the value if higher for example: $$f(8)>f(9)\Rightarrow I_2=[\frac{8+9}{2},9]=[\frac{17}{2},9]$$ and so on for $$J_n$$ as well. We end up with two successions of intervals $$I_n$$ and $$J_n$$ which bound eachother, one from the top and one from the bottom, the length of both intervals tends to $$\frac{1}{2^{n-1}}$$ which is infinitesimal. Both of these two intervals converge to $$9$$ and since the minimum of $$f(x)$$ is clearly between $$I_n$$ and $$J_n$$ we can conclude that $$9$$ is the minimum.
NOTE: This proof is strongly influenced by knowing beforehand that the minimum is $$9$$ through evaluating the derivative, there probably is a method that can find it on it's own but i couldn't come up with one. Either case this proof never uses the derivative, it only asks to evaluate $$f(x)$$ a couple of times in order to converge to the minimum.