Inverse of $f(x)=x^3-3x^2+4$ for all $x\geq2$

Question

I am trying to find an inverse of th function $$f(x)=x^3-3x^2+4$$for all $x\geq2$

it is easy to show that $$f(x)=x^3-3x^2+4=(x+1)(x-2)^2$$

I'm stuck here. Can I get some help please? Thanks!

Answer 1

The function is $0$ at $2$. The derivative of it is $3x^2-6x=3x(x-2)$ is $>0$ for $x>2$, hence the function is increasing from $0$ to $\infty$ and has an inverse $[2,\infty)\to [0,\infty)$.

The inverse function takes $x$ to the solution of the equation $y^3-3y^2+4=x$. You can use Cardano's formula, to find the solution. Note, that this equation has only one (positive) real solution for every $x\ge 2$, https://en.wikipedia.org/wiki/Cubic_equation#General_cubic_formula

$$y = 1 + 2^{1/3}/(-2 + x + \sqrt{-4 x + x^2})^{1/3} + (-2 + x + \sqrt{-4 x + x^2})^{1/3}/2^{1/3}.$$

Answer 2

I shall follow the steps given here.

Consider the cubic equation $$x^3-3x^2+(4-y)=0$$

The discriminant $$\Delta=-27 (y-4) y$$ is negative as soon as $y>4$ that is to say when $x>2$.

Using the hyperbolic method for one real root, then

$$x=1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{y}{2}-1\right)\right)$$