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I'm trying reverse the following equation, consider this:

enter image description here

i've done this to get a value to scale from Max to Min, example i put a x value bigger then i get a small value, so i've tried reverse and i've done something like this:

enter image description here

but i have not got back to the original value, what i'm doing wrong?

MindLerp
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  • I don't understand the question. Are you looking for the inverse of the function $f(x)=\frac {\sqrt x}x$? – lulu Sep 04 '17 at 23:36
  • Note: if that is what you are asking, note that $f(x)=\frac 1{\sqrt x}$. – lulu Sep 04 '17 at 23:38
  • @lulu exactly, I want to come back to the value obtained by that formula... – MindLerp Sep 04 '17 at 23:38
  • Ok, well then note that your function simplifies to $\frac 1{\sqrt x}$. Can you see the inverse now? – lulu Sep 04 '17 at 23:39
  • @lulu allright, you do not understand me, i do example: (sqrt(40) / 40) and i get 0.158114, and you suggest that I should do: (1 / sqrt(40)) for reverse it ? – MindLerp Sep 04 '17 at 23:46
  • No...Your function is $f(x)=\frac 1{\sqrt x}$. I mean that this is the same function as $\frac {\sqrt x}{x}$. I am pointing it out because it makes it obvious what the inverse function should be. – lulu Sep 04 '17 at 23:52
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    Just to be clear: if $y=\frac 1{\sqrt x}$ then $x=\frac 1{y^2}$. Thus the inverse function you seek is $f^{-1}(x)=\frac 1{x^2}$. – lulu Sep 04 '17 at 23:53
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    To use your example, $f(40)=\frac 1{\sqrt {40}}=0.158113883$. If we then apply $f^{-1}$ we get $f^{-1}(0.158113883)=\frac 1{(0.158113883)^2}=40$ as desired. – lulu Sep 04 '17 at 23:55
  • @lulu i'm so silly, sorry, i had forgotten that root to -1 it's the inverse to x value...thanks so much... – MindLerp Sep 05 '17 at 00:03
  • Oh, no problem at all. Good luck. – lulu Sep 05 '17 at 00:06

1 Answers1

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If you want to undo the original function, let $f(x)=v=\displaystyle\frac{\sqrt x}{x}$, then, to calculate $f^{-1}(x)$, all that needs to be done is you need to solve for the following $x$:

$v=\displaystyle\frac{\sqrt{x}}{x}$

square both sides

$v^2=\displaystyle\frac{1}{x}$

attain the reciprocal

$\displaystyle\frac{1}{v^2}=x$

and then swap the variables

$\displaystyle\frac{1}{x^2}=v$

hence, $f^{-1}(x)=\displaystyle\frac{1}{x^2}$

To confirm whether or not what we have written is true, we can take the rule that $f(f^{-1}(x))=x$, so lets substitute in $x=2$.

$f^{-1}(2)=\displaystyle\frac{1}{2^2}=\frac{1}{4}$

$\displaystyle f(\frac{1}{4})=\frac{\sqrt\frac{1}{4}}{\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{1}{4}}=\frac{1}{2}\cdot4=2$

We can go further to prove that in this case $f(f^{-1}(x))=x$ if we were to use induction on $x$.

joshuaheckroodt
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