inverse function Q (1 Viewer)

[mojako]

f(x): y = 3x - x^3
inverse f(x): x = 3y - y^3
can we make y the subject in inverse f(x)

Thanks.

EDIT: I was told this Q by someone, and it turns out this someone lied to me.. u dont need to be able to do this Q

 

[ngai]

use the cubic formula

 

[mojako]

grr im alergic to that formula

thanks ngai

 

[speersy]

what is the cubic formula?
is it ext 2?

 

[mojako]

yea.. what is it?
I thought I could use (A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3
but doesnt look like I can

if u mean something like
x= [-b +- sqrt(b^2 - 4ac)] / 2a
I must say the cubic one is really ugly

 

[ngai]

http://mathworld.wolfram.com/CubicFormula.html

 

[CrashOveride]

ngai seems to be pulling his usual inspection bash

i cant see it can be done nicely ??

 

[CrashOveride]

Also when we take derivative of f(x) and derivative of its inverse and multiply them, we get one ? What's the proof for this?

 

[Premus]

So how do you find the inverse function?
Im not sure how you use the Cubic formula!

Thanks

 

[mojako]

you wont need the cubic formula in any HSC mathematics courses.. inc ext2
(hmm.. maybe u might be asked to derive it with some guidelines haha)
its just a formula that ngai came across

this Q is from HSC 1994 so if u have the soln (MANSW or sumthing) can u tell us?
Thanks.

EDIT: no it's not from HSC 1994

 

[jumb]

If you have to graph this, it would be easier drawing it and flipping over the x=y line.

 

[nit]

Meh...Cardan Polynomials...brings back bad memories of a JL Williams comp a couple of years ago.

 

[Slidey]

f(x): y = 3x - x^3
inverse f(x): x = 3y - y^3
can we make y the subject in inverse f(x)

Thanks.

Well, since the cubic is already in the form y^3+Ay=B, there is no need to depress it.

So, that makes things slightly, er... simpler.

Anyway, I worked out that the 3 roots of the cubic are:

x=[(y+sqrt(y^2-4))/2]^(1/3)+[(y-sqrt(y^2-4))/2]^(1/3)
x=-0.5*{[(y+sqrt(y^2-4))/2]^(1/3)+[(y-sqrt(y^2-4))/2]^(1/3)} + 0.5i*sqrt(3)*{[(y+sqrt(y^2-4))/2]^(1/3)-[(y-sqrt(y^2-4))/2]^(1/3)}
x=-0.5*{[(y+sqrt(y^2-4))/2]^(1/3)+[(y-sqrt(y^2-4))/2]^(1/3)} - 0.5i*sqrt(3)*{[(y+sqrt(y^2-4))/2]^(1/3)-[(y-sqrt(y^2-4))/2]^(1/3)}

.'., inverse f(x) is:
[(x+sqrt(x^2-4))/2]^(1/3)+[(x-sqrt(x^2-4))/2]^(1/3)
or
-0.5*{[(x+sqrt(x^2-4))/2]^(1/3)+[(x-sqrt(x^2-4))/2]^(1/3)} + 0.5i*sqrt(3)*{[(x+sqrt(x^2-4))/2]^(1/3)-[(x-sqrt(x^2-4))/2]^(1/3)}
or
-0.5*{[(x+sqrt(x^2-4))/2]^(1/3)+[(x-sqrt(x^2-4))/2]^(1/3)} - 0.5i*sqrt(3)*{[(x+sqrt(x^2-4))/2]^(1/3)-[(x-sqrt(y^2-4))/2]^(1/3)}

Or maybe just the non-complex root. Yea... use that.

Although suffice it to say, you needn't worry about getting a question like this in an exam.

Oww my head.

 

[Slidey]

Hmm. I graphed that, and it actually DOES describe the inverse, up to the points where the stationary points and inflexion points come in, in which case perhaps you're meant to use the two compelx roots... maybe the inverse is a piecemeal function?

 

[CrashOveride]

Also when we take derivative of f(x) and derivative of its inverse and multiply them, we get one ? What's the proof for this?

No the question is not from HSC 1994, its similar to it but they did not require u to actually find the inverse. I was just wondering. They wanted you to find the gradient of the inverse function at x=0.