Find the inverse of the given function

Question

Let $$ f(x) = 4x^3 - 3x$$ Then find the inverse of the function for all admissible $x$.

I substituted $x= \cos t$ and then easily found out the inverse function but this is not applicable for all values of $x$. How can I find the result in general?

Answer 1

$$4\cos^3t-3\cos t=\cos 3t$$

and

$$4\cosh^3t-3\cosh t=\cosh 3t.$$

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The second identity yields

$$x=\cosh t=\cosh\left(\frac13\text{arcosh } y\right)=\cosh\left(\frac13\log\left(y+\sqrt{y^2-1}\right)\right)=\frac12\left(\sqrt[3]{y+\sqrt{y^2-1}}+\sqrt[3]{y-\sqrt{y^2-1}}\right).$$

This is compatible with Cardano's formula. And with $|y|<1$,

$$\frac12\left(\sqrt[3]{y+i\sqrt{1-y^2}}+\sqrt[3]{y-i\sqrt{1-y^2}}\right).$$

Answer 2

So we want to find a formula for $y$ such that $x=4y^3-3y$ so that means that $0=4y^3-3y-x$ and now we can apply the formula for finding the zeros of cubics. This formula is monstrous and ugly but it should work. This is what the other poster refers to as Cardano's formula. You can find it here: https://math.vanderbilt.edu/schectex/courses/cubic/. I find this tedious so I will use software but inspection of the formula makes the output of wolfram alpha seem reasonable.

Plugging "4y^3-3y-x zeros" into wolfram alpha yields:

$y = \frac{1}{2}[ {(\sqrt{x^2 - 1} + x)}^{1/3} + {(\sqrt{x^2 - 1} + x)}^{-1/3}]$

Note that this is not defined when $|x|<1$. But we need to exclude $|x|=1$ as well: If $f(x)=4x^3-3x$ then $1=4x^3-3x$ implies that $x=-\frac{1}{2}$ or $x=1$ so we need to exclude this from our inverse. So for $|x|>1$ we have $f^{-1}(x)= \frac{1}{2}[ {(\sqrt{x^2 - 1} + x)}^{1/3} + {(\sqrt{x^2 - 1} + x)}^{-1/3}]$.