How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$

Question

It is a question from a quiz.

The following is the whole question.

Let \begin{eqnarray} \\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0), \end{eqnarray} find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.

The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form.

I need help.

Update : Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?

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Thank you for your attention

Answer 1

Hint: Try matching first and last terms: $A^3 = x^3$ and $B^3=-\frac{64}{x^3}$ and check if it fits the other terms.

Answer 2

Hint: expand $$\left(x - \frac 4x\right)^3$$

Answer 3

Hint: $y = \left(x - \dfrac4x\right)^3 \Longrightarrow x^2 - x\sqrt[3]{y} - 4 = 0$.

Answer 4

for $x \in (-\infty,0)$ the inverse function has the form

$$ f^{-1}(x)=\frac{1}{2}\left( \sqrt[3]{x}-\sqrt{\sqrt[3]{x^2}+16} \right) $$ As mentioned above we have

$y = \left(x - \dfrac4x\right)^3 \Longrightarrow x^2 - x\sqrt[3]{y} - 4 = 0$.

By solving the equation we get $$ x=\frac{1}{2}\left( \sqrt[3]{y}-\sqrt{\sqrt[3]{y^2}+16} \right), $$ and $$ x=\frac{1}{2}\left( \sqrt[3]{y}+\sqrt{\sqrt[3]{y^2}+16} \right). $$ The first expression is always negative (note $x \in (-\infty,0)$ but second is positive and not suitable for us. Thus, changing the variables, we get the inverse function.

Answer 5

$\left(x-\frac{4}{x}\right)^{3}$

Answer 6

First of all, $f$ does not have a inverse function because you always have two root of $f - y$. One positive root, one negative root.

If you add the constraints that $x$ is positive and $y$ is real, then there will be an inverse function. $$(x - \frac4x)^3=y$$

Since $y$ is real:

$$x - \frac4x=\sqrt[3]y$$

Since $x$ is not zero.

$$x^2 - 4=\sqrt[3]yx$$

Solve the quadratic equation:

$$x^2 - \sqrt[3]yx = 4 $$ $$(x-\frac{\sqrt[3]y}2)^2=4+(\frac{\sqrt[3]y}2)^2$$ $$ x - \frac{\sqrt[3]y}2 = \pm \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$ $$ x = \frac{\sqrt[3]y}2 \pm \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$

Since $x$ is positive, and the magnitude of the second term is greater than the first term, so you have to take the plus sign. This is your inverse function: $$f^{-1}(y) = \frac{\sqrt[3]y}2 + \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$

If, however, you constrain $x$ to be negative, then take the minus sign above.

Answer 7

I first would eliminate the negative exponents of $x$ by multiplying $f(x)$ by $x^3$ to get $$g(x)=x^3f(x)=x^6-12\,x^4+48\,x^2-64$$ If you are right in assuming that $f(x)=h(x)^3$ then $g(x)=p(x)^3$ and $g'(x)=3p'(x)p(x)^2$ Now we assume that $p$ is a polynomial then $$p(x)^2 \mid \text{gcd}(g(x),g'(x))$$ and $$\frac{g(x)}{\text{gcd}(p(x),p'(x))} \mid p(x)$$

We get $$p'(x)=6\,x^5-48\,x^3+96\,x$$ and $$\text{gcd}(p(x),p'(x))=x^4-8\,x^2+16$$

$$\frac{g(x)}{\text{gcd}(p(x),p'(x))}=x^2-4=(x-2)(x+2)$$ Because the degree of the last polynomial is $2$ we must have $$p(x)=x^2-4$$ (ignoring constant factors) We can check that $$x^6-12\,x^4+48\,x^2-64=(x^2-4)^3=(x-2)^3(x+2)^3$$ and so $$f(x)=(x-\frac{4}{x})^3=(\sqrt{x}-\frac{2}{\sqrt{x}})^3 (\sqrt{x}+\frac{2}{\sqrt{x}})^3$$

Now let's solve $$(x-2/x)=y$$ for $x$. By multiplying the equation by $x<$ we finally get the quadratic equation $$x^2-yx-2=0$$ and get

$$x=-{{\sqrt{y^2+8}-y}\over{2}} $$ $$x=+{{\sqrt{y^2+8}+y}\over{2}}$$

So we have two inverse images for each value $y$.