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i am finding a inverse function to $$y=\frac{e^\frac{x^3}{3}}{x^3}$$ First step: $$x=\frac{e^\frac{y^3}{3}}{y^3}$$ Next: ... $$(xy^3)^3 = e^{{y}^3}$$

and now in this step i have no idea how can i express "y"

$$\ln(x)+3\ln(y) = y$$

egreg
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DavidM
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    are you sure that an inverse function exists? – Dr. Sonnhard Graubner Nov 13 '15 at 18:24
  • i am not sure.. i would like to finde Range of this funciton, is it a good way finding inverse funciton? – DavidM Nov 13 '15 at 18:27
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    Where did you get $\ln(x)+3\ln(y)$? If you take the log of both sides, you get $3\ln x + 9\ln y = y^3$. In any event, this sort of function generally doesn't have an inverse without the Lambert $W$-function. – Thomas Andrews Nov 13 '15 at 18:30
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    This is a classic example of the XY problem. You really need to find the range, so your attempt is to find the inverse function which is reasonable. However this isn't the best way. If your problem is to find the range, ask to find the range. –  Nov 13 '15 at 18:30
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    http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem –  Nov 13 '15 at 18:32

2 Answers2

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Go back to the original:

$$y = x^{-3}e^{x^3/3}$$

Then: $$\frac{-1}{y}=-x^3e^{-x^3/3}$$ So: $$\frac{-1}{3y} = \frac{-x^3}{3}e^{-x^3/3}$$ This means that:

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

where $W$ is Lambert's $W$-function, with $W$ an inverse of $ze^z$.

So $$x=\left(-3W\left(\frac{-1}{3y}\right)\right)^{1/3}$$

You are not going to be able to do it without Lambert or some other non-elementary function.

Thomas Andrews
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Consider the function $$ f(x)=\frac{e^{x^3/3}}{x^3} $$ This function is defined for $x\ne0$. Also $$ \lim_{x\to-\infty}f(x)=0,\quad \lim_{x\to0^-}f(x)=-\infty,\quad \lim_{x\to0^+}f(x)=\infty,\quad \lim_{x\to\infty}f(x)=\infty $$ so the function has no “global” inverse.

Its derivative is $$ f'(x)=\frac{x^5e^{x^3/3}-3x^2e^{x^3/3}}{x^6}=\frac{x^3-3}{x^4}e^{x^3/3} $$ which vanishes for $x=\sqrt[3]{3}$ and is negative for $x<\sqrt[3]{3}$ (but $x\ne0$, of course) and positive for $x>\sqrt[3]{3}$.

So the function is decreasing in the intervals $(-\infty,0)$ and $(0,\sqrt[3]{3}]$; it is increasing in the interval $[\sqrt[3]{3},\infty)$.

How can we find the range? We know that the interval $(-\infty,0)$ is contained in the range (consider the $(-\infty,0)$ part of the domain). Since $$ f(\sqrt[3]{3})=\frac{e}{3} $$ we know that the range corresponding to the part $(0,\infty)$ of the domain is $[e/3,\infty)$.

Putting things together, the range is $$ (-\infty,0)\cup[e/3,\infty) $$

The restriction of $f$ to each of those intervals has an inverse function. Which is, well, the inverse function: there's no “explicit” form for it. But it's not necessary to find an inverse function, in order to determine the range.

egreg
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