What is the Minimum of f(x) Without Using Calculus?

[anemone]

Without using the calculus, determine the minimum of $f(x)=(x^2-6)(x^2-8)(x^2-10)(x^2-12)+(6\cdot 8\cdot 10\cdot 12)$.

 

[kaliprasad]

Without using the calculus, determine the minimum of $f(x)=(x^2-6)(x^2-8)(x^2-10)(x^2-12)+(6\cdot 8\cdot 10\cdot 12)$.

Spoiler

given expression = $(x^2-6)(x^2-12)(x^2-8)(x^2-10) + 6\cdot 8 \cdot 10\cdot 12 $
=$(x^4-18x^2 + 72)(x^4-18x^2 + 80) +6\cdot 8 \cdot 10\cdot 12 $
=$((x^2-9)^2 - 9) ((x^2-9)^2 - 1) +6\cdot 8 \cdot 10\cdot 12 $
the 2nd term is constant and the 1st term is product of 2 terms

for $(x^2-9)$ less than 1 or greater than 9 both terms are positive
for it between 1 and 9 one term is positive and another is -ve and is lowest when $x^2-9$ in the middle that is 5 and lowest value becomes 5744

 

[anemone]

Thanks for participating in this challenge, kaliprasad!

My solution, which the method used is more or less the same as yours:

Spoiler

Let $y=x^2-9$, the original equation becomes

$\begin{align*}f(y)&=(y+3)(y+1)(y-1)(y-3)+6\cdot 8\cdot 10\cdot 12\\&=(y^2-9)(y^2-1)+6\cdot 8\cdot 10\cdot 12\\&=y^4-10y^2+9+6\cdot 8\cdot 10\cdot 12\\&=(y^2-5)^2+(6\cdot 8\cdot 10\cdot 12-16)\end{align*}$

Now, there are two ways to find the minimum of $f(x)$:

1. We can actually conclude, at this point the the minimum of $f(x)=6\cdot 8\cdot 10\cdot 12-16=5744$.

2. We can conclude first up to this point that the minimum of $f(y)$ occurs at $y^2=5,\,\rightarrow y=\pm \sqrt{5}$ and substituting this back to $f(x)$ to get the minimum of the function of $f(x)$, we get

$\begin{align*}f(x)_{\text{minimum}}&=(\pm\sqrt{5}-9+6)(\pm\sqrt{5}-9+8)(\pm\sqrt{5}-9+10)(\pm\sqrt{5}-9+12)+6\cdot 8\cdot 10\cdot 12\\&=(\pm\sqrt{5}-3)(\pm\sqrt{5}+3)(\pm\sqrt{5}-1)(\pm\sqrt{5}+1)+6\cdot 8\cdot 10\cdot 12\\&=(5-9)(5-1)+6\cdot 8\cdot 10\cdot 12\\&=-16+5760\\&=5744\end{align*}$

 

[kaliprasad]

Thanks for participating in this challenge, kaliprasad!

My solution, which the method used is more or less the same as yours:

Spoiler

Let $y=x^2-9$, the original equation becomes

$\begin{align*}f(y)&=(y+3)(y+1)(y-1)(y-3)+6\cdot 8\cdot 10\cdot 12\\&=(y^2-9)(y^2-1)+6\cdot 8\cdot 10\cdot 12\\&=y^4-10y^2+10+6\cdot 8\cdot 10\cdot 12\\&=(y^2-5)^2+(6\cdot 8\cdot 10\cdot 12-25)\end{align*}$

We can conclude first up to this point that the minimum of $f(y)$ occurs at $y^2=5,\,\rightarrow y=\pm \sqrt{5}$ and substituting this back to $f(x)$ to get the minimum of the function of $f(x)$, we get

$\begin{align*}f(x)_{\text{minimum}}&=(\pm\sqrt{5}-9+6)(\pm\sqrt{5}-9+8)(\pm\sqrt{5}-9+10)(\pm\sqrt{5}-9+12)+6\cdot 8\cdot 10\cdot 12\\&=(\pm\sqrt{5}-3)(\pm\sqrt{5}+3)(\pm\sqrt{5}-1)(\pm\sqrt{5}+1)+6\cdot 8\cdot 10\cdot 12\\&=(5-9)(5-1)+6\cdot 8\cdot 10\cdot 12\\&=-16+5760\\&=5744\end{align*}$

so much calculation is not required as putting $y^2-5 = 0 $ we get the result directly

 

[anemone]

so much calculation is not required as putting $y^2-5 = 0 $ we get the result directly.

Yes, that was inefficient of me not to observe more because if I did, I would save some extra tedious work to figure out the minimum of $f$, and I could really just deduce it, just like you did. (Nerd)

 

[kaliprasad]

Thanks for participating in this challenge, kaliprasad!

My solution, which the method used is more or less the same as yours:

Spoiler

Let $y=x^2-9$, the original equation becomes

$\begin{align*}f(y)&=(y+3)(y+1)(y-1)(y-3)+6\cdot 8\cdot 10\cdot 12\\&=(y^2-9)(y^2-1)+6\cdot 8\cdot 10\cdot 12\\&=y^4-10y^2+10+6\cdot 8\cdot 10\cdot 12\\&=(y^2-5)^2+(6\cdot 8\cdot 10\cdot 12-25)\end{align*}$

We can conclude first up to this point that the minimum of $f(y)$ occurs at $y^2=5,\,\rightarrow y=\pm \sqrt{5}$ and substituting this back to $f(x)$ to get the minimum of the function of $f(x)$, we get

$\begin{align*}f(x)_{\text{minimum}}&=(\pm\sqrt{5}-9+6)(\pm\sqrt{5}-9+8)(\pm\sqrt{5}-9+10)(\pm\sqrt{5}-9+12)+6\cdot 8\cdot 10\cdot 12\\&=(\pm\sqrt{5}-3)(\pm\sqrt{5}+3)(\pm\sqrt{5}-1)(\pm\sqrt{5}+1)+6\cdot 8\cdot 10\cdot 12\\&=(5-9)(5-1)+6\cdot 8\cdot 10\cdot 12\\&=-16+5760\\&=5744\end{align*}$

the f(y) should be

$\begin{align*}f(y)&=(y+3)(y+1)(y-1)(y-3)+6\cdot 8\cdot 10\cdot 12\\&=(y^2-9)(y^2-1)+6\cdot 8\cdot 10\cdot 12\\&=y^4-10y^2+9+6\cdot 8\cdot 10\cdot 12\\&=(y^2-5)^2+(6\cdot 8\cdot 10\cdot 12-16)\end{align*}$

 

[anemone]

the f(y) should be

$\begin{align*}f(y)&=(y+3)(y+1)(y-1)(y-3)+6\cdot 8\cdot 10\cdot 12\\&=(y^2-9)(y^2-1)+6\cdot 8\cdot 10\cdot 12\\&=y^4-10y^2+9+6\cdot 8\cdot 10\cdot 12\\&=(y^2-5)^2+(6\cdot 8\cdot 10\cdot 12-16)\end{align*}$

Oh My...it seems I misunderstood of what you told me (implicitly) yesterday and that my typo led me to think I have to substitute the $y$ value back to get the minimum of $f(x)$...

I will fix the error now, thanks for your patient to make me realized what did I miss in this case.