Find the inverse function of $f(x) = \frac{x^3+3x}{2}$

My thought on this are as follows:

I've solved a similar problem where I had to find the inverse of:

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

So let $a = \sqrt[3]{x+\sqrt{x^2-1}}$ and $b = \sqrt[3]{x-\sqrt{x^2-1}}$, then $y = a + b$ and hence:

$$ y^3 = (a+b)^3 = a^3 + b^3 + 3ab(a+b) = \\ = 2x+3ab(a+b)=2x+3(a+b) $$

But $y = a + b$ and then

$$ y^3 = 2x + 3y \iff x = \frac{y^3-3y}{2} $$

This function is very similar to the one in the title.

The answer says the inverse $f^{-1}(x) = \sqrt[3]{x+\sqrt{x^2+1}} + \sqrt[3]{x-\sqrt{x^2+1}}, \; x\in \mathbb R$. So knowing the answer it's easy to show that it indeed appears to be the inverse of $f(x)$ using the approach above. But how could I achieve the same result given the inverse function is unknown?

  • 1
    $\begingroup$ You can solve a cubic equation. Here's a link $\endgroup$ – Jakobian Jul 7 '18 at 15:35
  • $\begingroup$ @Adam Will try this out, thanks $\endgroup$ – roman Jul 7 '18 at 15:38


Let $$x:=\sqrt[3]u-\frac1{\sqrt[3]u}$$ Then $$x^3+3x=u-3\sqrt[3]u+\frac3{\sqrt[3]u}-\frac1u+3\sqrt[3]u-\frac3{\sqrt[3]u}=u-\frac1u.$$

Now solve the equation


for $u$ (it can be reduced to a quadratic one).

In the end,


Notice that the product of the two $u$ roots is $-1$, so that the solution can also be expressed as the sum of the cubic roots.

  • $\begingroup$ Thanks, I've just managed to reach the same answer and got $x = \sqrt[3]{y\pm\sqrt{y^2+1}} +\sqrt[3]{y\mp\sqrt{y^2+1}} $ $\endgroup$ – roman Jul 7 '18 at 16:13

You want to solve $x^3 + 3x - 2y = 0$ as $x = f(y)$. The canonical way to do this is to write $x = a + b$, obtaining $$a^3 + 3a^2b + 3ab^2 + b^3 + 3(a + b) - 2y = 0$$ The trick is specify that $3ab = -3$. The above equation then reduces to $$a^3 - b^3 - 2y = 0$$ Which is the same as $$a^3 - 1/a^3 - 2y = 0$$ Multiplying by $a^3$ gives $$a^6 - 2ya^3 - 1 = 0$$ You can now use the quadratic formula to find an expression for $a^3$ in terms of $y$, and then you get three possible complex cube roots for $a$. You then can find $b$ through $3ab = -3$ and then you have three possible $x = a + b$. These are the solutions to your original equation. The simplest of these three will be given by the formula you have there.


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