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

[Bardagath]

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

 

[Mark44]

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.

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.

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.

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.

 

[Bardagath]

Thanks so much,

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

Thankyou very much