# What is the inverse of the function

1. May 3, 2006

### kmarinas86

I'm running out of ideas:

$$x=ay+by^3$$

Does someone here now how to solve for $$y$$ in terms of $$x$$?

Last edited: May 3, 2006
2. May 3, 2006

### Curious3141

$$y^3 + \frac{a}{b}y - \frac{x}{b} = 0$$

Two ways :

1) Substitute $$y = z - \frac{a}{3bz}$$. Form a quadratic in $$z^3$$, solve for z and find y.

2) Compare equation to the trig identity $$cos^3 \theta - \frac{3}{4}\cos\theta - \frac{1}{4}\cos 3\theta = 0$$ while letting $$y = m\cos\theta$$ then comparing coefficients. With this method, if you have to compute the arccosine of a value greater than one in magnitude, use the identity $$\cos i\theta = \cosh \theta$$

These are methods used to solve the general cubic in radicals/trig/hyperbolic trig ratios when the second degree term is missing (or has been eliminated).

Last edited: May 3, 2006
3. May 9, 2006

### kmarinas86

Last edited: May 9, 2006