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

Question

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...

Answer 1

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 $(9,243)$ .

Answer 2

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 $y_{\mathrm{min}} = 243$ .

Answer 3

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.

Answer 4

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 $0$ .

Answer 5

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. $x= 9$ .

Answer 6

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.

Answer 7

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

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