anonymous
  • anonymous
help with verifying an inverse
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
so i have \[f(x)=\frac{ 100 }{ 1+2^{-x} }\] and \[f ^{-1}(x)=-\log_{2}(\frac{ 100 }{ x } -1)-1\]
anonymous
  • anonymous
I need help with \[(f o f ^{-1})x=x\]
anonymous
  • anonymous
because after I distribute the negative I am not sure what to do with the log in the exponent of 2

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anonymous
  • anonymous
so I'm stuck on\[\frac{ 100 }{ 1+2^{\log_{2}(\frac{ 100 }{ x }-1)+1 } }\]
Luigi0210
  • Luigi0210
Oh, I got this!
Luigi0210
  • Luigi0210
Do you know your log rules?
anonymous
  • anonymous
yes
Luigi0210
  • Luigi0210
Use PEMDAS
anonymous
  • anonymous
but I don't understand how to progress is my problem like because it is 2^logbase 2 does that just mean that all the other stuff drops into the bottom
anonymous
  • anonymous
so I would have\[\frac{ 100 }{ 1-1+\frac{ 100 }{ x }-1+1 }\]
Luigi0210
  • Luigi0210
Well if it works then I suppose
anonymous
  • anonymous
but it doesnt work which is what I dont understand because for me (f^-1 o f)(x) worked but not when I plug f into f^-1
anonymous
  • anonymous
\[ f(x) = \frac{100}{1+2^{-x}}\\ f^{-1}(x) = -\log_2\left(\frac{100}{x}-1\right)-1 \] \[ \begin{align*}\left(f\circ f^{-1}\right)(x)&=f\left(f^{-1}(x)\right)\\\\ &= \dfrac{100}{1+2^{-\left[-\log_2\left(\dfrac{100}{x}-1\right)-1\right]}}\\\\ &=\dfrac{100}{1+2^{\log_2\left(\dfrac{100}{x}-1\right)+1}}\\\\ &=\dfrac{100}{1+2\cdot2^{\log_2\left(\dfrac{100}{x}-1\right)}}\\\\ &=\dfrac{100}{1+2\left(\dfrac{100}{x}-1\right)}\\\\ &=\dfrac{100}{1+\dfrac{200}{x}-2}\\\\ &\not=x\end{align*} \] This might amount to a typo. Did you mean \(f^{-1}(x)=-\log_2\left(\dfrac{100}{x}-1\right)\) ?
anonymous
  • anonymous
Since http://www.wolframalpha.com/input/?i=inverse+of+100%2F%281%2B2%5E%28-x%29%29
anonymous
  • anonymous
wait so how would (100/x-1) be any better if you get 200/x so that doesnt cancel
anonymous
  • anonymous
That's what I'm saying... Are you sure that's the right \(f^{-1}(x)\) you've provided? Because it doesn't work.
anonymous
  • anonymous
well no I mean the new one you gave with it being just 100/x-1 instead of 100/x-1-1 because is it still multiplied by two based on what you showed before?
anonymous
  • anonymous
because when you were disproving my f^-1 i do not understand what you did with the -1 outside the paranthesis
anonymous
  • anonymous
So between the third and fourth lines of work?
anonymous
  • anonymous
yes
anonymous
  • anonymous
I used a property of exponents. You see how \(2^3=2^{2+1}\) ? Well a property of exponents is that the base is "distributed," like so: \(2^{2+1}=2^2\cdot2^1\)
anonymous
  • anonymous
oh ok i see so if i got rid of the one outside the paraenthesis it should all work out because you're right I see the error that gave me the wrong f^-1
anonymous
  • anonymous
hey man I highly appreciate all your help and for coming back after waiting an hour while I was at cross country practice
anonymous
  • anonymous
I wasn't exactly waiting, but you're welcome all the same.
anonymous
  • anonymous
well just being there I mean I still highly appreciate it
anonymous
  • anonymous
happy to help

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