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March 7th, 2017, 02:31 AM   #1
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inverting polynomial

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My efforts to find an inverse for this equation: f(x) = (3x^3 + 6x^2 + 7x), exists or not are in vain. Could anyone help me try to solve this. I would be grateful.
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March 7th, 2017, 12:29 PM   #2
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$f(x) = 3x^3 + 6x^2 + 7x$

$f'(x) = 9x^2 + 12x + 7$

Functions with inverses must be monotonic over their natural or restricted domain.

Is $f(x)$ monotonic on $\mathbb{R}$?

How might you tell?
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March 7th, 2017, 06:17 PM   #3
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Actually the f(x) is over Zn where n = 9. So the problem is to find if f(x) mod 9 has an inverse or not.
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March 7th, 2017, 06:24 PM   #4
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Quote:
Originally Posted by classkid View Post
Actually the f(x) is over Zn where n = 9. So the problem is to find if f(x) mod 9 has an inverse or not.
That little piece of info should have been in the first post.

You folks just love to waste our time.
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March 8th, 2017, 02:49 AM   #5
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I am sorry. But could you please help me solving this problem?
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March 8th, 2017, 09:06 AM   #6
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The following table of values of the 9 elements mod 9 shows that the function is indeed both surjective and injective. Hence it does have an inverse.

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March 8th, 2017, 11:05 AM   #7
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3x² + x
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March 8th, 2017, 06:31 PM   #8
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So if I want to find its inverse in an equation form, what would be my approach? Could u please tell me.
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March 8th, 2017, 11:34 PM   #9
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Quote:
Originally Posted by skipjack View Post
3x² + x
I'd be interested in knowing how you came up with this.

It's not unique. There are two other "inverses".

$3x^3 + 3x^2 + 7x$

$6x^3 + 3x^2 + 4x$

I determined these through the most brutish of force.
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March 9th, 2017, 12:37 AM   #10
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I just looked for a quadratic and found one.

Once a solution is known, others follow as 3x(x - 1)(x + 1) ≡ 0 (mod 9).
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