Question

help with verifying an inverse

Answer 1

so i have \[f(x)=\frac{ 100 }{ 1+2^{-x} }\] and \[f ^{-1}(x)=-\log_{2}(\frac{ 100 }{ x } -1)-1\]

Answer 2

I need help with \[(f o f ^{-1})x=x\]

Answer 3

because after I distribute the negative I am not sure what to do with the log in the exponent of 2

Answer 4

so I'm stuck on\[\frac{ 100 }{ 1+2^{\log_{2}(\frac{ 100 }{ x }-1)+1 } }\]

Answer 5

Oh, I got this!

Answer 6

Do you know your log rules?

Answer 7

yes

Answer 8

Use PEMDAS

Answer 9

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

Answer 10

so I would have\[\frac{ 100 }{ 1-1+\frac{ 100 }{ x }-1+1 }\]

Answer 11

Well if it works then I suppose

Answer 12

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

Answer 13

\[ f(x) = \frac{100}{1+2^{-x}}\\
f^{-1}(x) = -\log_2\left(\frac{100}{x}-1\right)-1 \]
\[ \begin{align*}\left(f\circ f^{-1}\right)(x)&=f\left(f^{-1}(x)\right)\\\\
&= \dfrac{100}{1+2^{-\left[-\log_2\left(\dfrac{100}{x}-1\right)-1\right]}}\\\\
&=\dfrac{100}{1+2^{\log_2\left(\dfrac{100}{x}-1\right)+1}}\\\\
&=\dfrac{100}{1+2\cdot2^{\log_2\left(\dfrac{100}{x}-1\right)}}\\\\
&=\dfrac{100}{1+2\left(\dfrac{100}{x}-1\right)}\\\\
&=\dfrac{100}{1+\dfrac{200}{x}-2}\\\\
&\not=x\end{align*} \]
This might amount to a typo. Did you mean \(f^{-1}(x)=-\log_2\left(\dfrac{100}{x}-1\right)\) ?

Answer 14

Since http://www.wolframalpha.com/input/?i=inverse+of+100%2F%281%2B2%5E%28-x%29%29

Answer 15

wait so how would (100/x-1) be any better if you get 200/x so that doesnt cancel

Answer 16

That's what I'm saying... Are you sure that's the right \(f^{-1}(x)\) you've provided? Because it doesn't work.

Answer 17

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?

Answer 18

because when you were disproving my f^-1 i do not understand what you did with the -1 outside the paranthesis

Answer 19

So between the third and fourth lines of work?

Answer 20

yes

Answer 21

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: \(2^{2+1}=2^2\cdot2^1\)

Answer 22

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

Answer 23

hey man I highly appreciate all your help and for coming back after waiting an hour while I was at cross country practice

Answer 24

I wasn't exactly waiting, but you're welcome all the same.

Answer 25

well just being there I mean I still highly appreciate it

Answer 26

happy to help