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i am finding range of this function

$$y=\frac{e^\frac{x^3}{3}}{x^3}$$

As a first step i tried find inverse function, but i cannot find it. Please can you help me with it?

Thomas Andrews
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DavidM
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  • Is there a reason the $x$- and $y$-coordinates seem swapped? This looks like $x=f(y)$ which is usually the opposite of standard notation. – Addem Nov 13 '15 at 18:46
  • sorry it was misstake – DavidM Nov 13 '15 at 18:46
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    http://math.stackexchange.com/questions/1527569/express-y-from-lnx3-lny-y/1527606#1527606 – DavidM Nov 13 '15 at 18:47
  • OK, final question before attempting an answer: In what context is the question asked? Algebra, Calculus, or otherwise? The best method of solving may depend on the answer. – Addem Nov 13 '15 at 18:48
  • i have a function and i need find Domain, Range, interval where it is continues, extrems, convexity, concavity... – DavidM Nov 13 '15 at 18:51
  • Finding the "inverse function" is much harder than finding the range of the function. – GEdgar Nov 13 '15 at 18:51
  • Why would you look for an inverse function when all you need to calculate is the range? – 5xum Nov 13 '15 at 18:51

1 Answers1

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As noted in the other question related to this function:

$$-\frac{1}{3y} = ue^{u}$$ where $u=-x^3/3.$ The range of $ue^u$ is well-known when $u$ is real. We have to exclude $u=0$, since the original function is undefined when $x=0$.

Now $x\to -x^3/3$ is invertible on the real line, so you just need to find the range of $ue^u$, exclude $u=0$, invert, negate, and divide by $3$.

$ue^u$ has range equal to $(-1,+\infty)$. Applying that transformation, inverting we get $(-\infty,-1)\cup (0,+\infty)$, negating, we get $(-\infty,0)\cup (1,+\infty)$, and dividing by $3$ we get:

$$(-\infty,0)\cup (1/3,+\infty)$$

Thomas Andrews
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