Solve Inverse Function: Find f^-1(x) with y=6x^3+6x+2

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SUMMARY

The discussion focuses on finding the inverse function of y=f(x)=6x^3+6x+2. The initial transformation x=6y^3+6y+2 leads to the equation x-2=6y^3+6y, which is a cubic polynomial. The solution involves defining new variables u and v, where uv=2 and x=u-v, followed by substituting v=2/u into the function. This results in a quadratic equation in terms of u^3, which can be solved to express x in terms of y.

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shiri
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I got a problem for this question.

Given the function y=f(x)=6x^3+6x+2. Find f^-1(x).

Can anybody tell me how to solve this function?So far I got this:

x=6y^3+6y+2

x-2=6y^3+6yI just couldn't solve for y variable. So, please tell what I do wrong?

thanks
 
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It doesn't look like you are doing anything wrong.
Just note that it is a cubic polynomial. So although
x-2 = 6y3+6y
will definitely have solutions for y(x), they may be non-trivial to write down.

However, a solution exists.
 
shiri said:
I got a problem for this question.

Given the function y=f(x)=6x^3+6x+2. Find f^-1(x).

Can anybody tell me how to solve this function?


So far I got this:

x=6y^3+6y+2

x-2=6y^3+6y


I just couldn't solve for y variable. So, please tell what I do wrong?

thanks

You couldn't because it is not that much easy. I'm going to sort some steps that you should carefully follow to get x in terms of y:

1- Define two new variables u and v, with uv = 2 and x=u-v. Introduce this into your function and simplify the result.

2- Put v=2/u in the new function.

3- Now you have a simple function which is quadratic in u^3. From there, I assume, you know how to solve a quadratic equation for t=u^3 after taking y to the right side and letting one side just be lonely with a zero!

4- Now from 1,2 and 3 we have [tex]x=u-2/u = t^{1/3}-2t^{-1/3}[/tex] and this is what you want since t is a function of y.

AB
 

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