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

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The discussion focuses on finding the minimum of the function f(x) = (x^2-6)(x^2-8)(x^2-10)(x^2-12) + (6·8·10·12) without using calculus. Participants share their methods, noting that substituting y^2 - 5 = 0 simplifies the process significantly. There is acknowledgment of inefficiencies in previous calculations, with one participant realizing they misunderstood the approach needed to find the minimum. The conversation emphasizes the importance of recognizing simpler methods to solve the problem effectively. Overall, the thread highlights collaborative problem-solving and learning from mistakes.
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Without using the calculus, determine the minimum of $f(x)=(x^2-6)(x^2-8)(x^2-10)(x^2-12)+(6\cdot 8\cdot 10\cdot 12)$.
 
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anemone said:
Without using the calculus, determine the minimum of $f(x)=(x^2-6)(x^2-8)(x^2-10)(x^2-12)+(6\cdot 8\cdot 10\cdot 12)$.
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
 
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Thanks for participating in this challenge, kaliprasad!

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

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

$\begin{align*}f(x)_{\text{minimum}}&=(\pm\sqrt{5}-9+6)(\pm\sqrt{5}-9+8)(\pm\sqrt{5}-9+10)(\pm\sqrt{5}-9+12)+6\cdot 8\cdot 10\cdot 12\\&=(\pm\sqrt{5}-3)(\pm\sqrt{5}+3)(\pm\sqrt{5}-1)(\pm\sqrt{5}+1)+6\cdot 8\cdot 10\cdot 12\\&=(5-9)(5-1)+6\cdot 8\cdot 10\cdot 12\\&=-16+5760\\&=5744\end{align*}$
 
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anemone said:
Thanks for participating in this challenge, kaliprasad!

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

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
 
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kaliprasad said:
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)
 
anemone said:
Thanks for participating in this challenge, kaliprasad!

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

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

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