How do I find the inverse of x^3/x^2+1?

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Bardagath
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Hello,

I have been working on this problem for a while now and can't seem to get it into a forum which would isolate x, which would then allow me to sub y and find the inverse.

The problem is x^3/x^2+1

Can anyone help me with this? I am utterly stuck

I have dealt with a similar problem before... x+1/x-1... the trick for that was to express the statement twice, once with x-1 in the numerator and x+2 in the numerator... which would cancel the first statement and isolate x etc.

However, this problem we have a cube and a square in the numerator and the denomator.

Can someone please help?

Thanks
 
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Bardagath said:
Hello,

I have been working on this problem for a while now and can't seem to get it into a forum which would isolate x, which would then allow me to sub y and find the inverse.

The problem is x^3/x^2+1
Presumably you mean f(x) = x3/(x2 + 1). What you wrote would be interpreted as (x3/x2) + 1 = x + 1.
Bardagath said:
Can anyone help me with this? I am utterly stuck

I have dealt with a similar problem before... x+1/x-1...
The similar problem was y = (x + 1)/(x - 1), I'm pretty sure.
Bardagath said:
the trick for that was to express the statement twice, once with x-1 in the numerator and x+2 in the numerator... which would cancel the first statement and isolate x etc.
(x + 1)/(x - 1) = (x - 1 + 2)/(x - 1) = (x - 1)/(x - 1) + 2/(x - 1) = 1 + 2/(x - 1). Is this what you're talking about? If so, it's fairly easy to solve the equation y = 1 + 2/(x - 1) for x.
Bardagath said:
However, this problem we have a cube and a square in the numerator and the denomator.

Can someone please help?

Thanks
Solving the equation y = x3/(x2 + 1) for x is possible, but pretty difficult.

Multiply both sides by x2 + 1:
y(x2 + 1) = x3

Expand the left side and bring all terms to the left side:
yx2 + y - x3 = 0

Rearrange by powers of x:
-x3 + yx2 + y = 0

Solving for x amounts to finding the solutions of this cubic equation, a technique that has been around for a long time, but isn't usually taught.
 
Thanks so much,

I was afraid I'd be left in this form. Oh well, must use Cardano's equation now...

Thankyou very much
 

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