+118667 This is not correct 
Can anyone spot what mistake I have made??
\(f(x)=\frac{2^x}{1+2^x}\\ let \\ y=\frac{2^x}{1+2^x}\\ \text{An initial inspection tells me that y>0}\\ \text{make x the subject}\\ y=\frac{2^x}{1+2^x}\\ \frac{1}{y}=\frac{1+2^x}{2^x}\\ \frac{1}{y}=\frac{1}{2^x}+\frac{2^x}{2^x}\\ \frac{1}{y}=\frac{1}{2^x}+1\\ \frac{1}{y}-1=2^{-x}\\ \frac{1-y}{y}=2^{-x}\\ log_2({\frac{1-y}{y}})=log_2(2^{-x})\\ log_2({\frac{1-y}{y}})=-x*log_2(2)\\ log_2({\frac{1-y}{y}})=-x\\ x=-log_2({\frac{1-y}{y}})\\ x=log_2(({\frac{1-y}{y}})^{-1})\\ x=-log_2\left({\frac{y}{1-y}}\right)\\ \therefore f^{-1}(x)=-log_2\left({\frac{x}{1-x}}\right)\\ \)
I have graphed this here
https://www.desmos.com/calculator/qt6xarwxbp
AND IT IS NOT CORRECT
Can anyone spot what mistake I have made??
+118667 Thanks Heureka,
I found my own error too :)
It was just a careless sign error in my last line.
\(x=log_2(({\frac{1-y}{y}})^{-1})\\ x=-log_2\left({\frac{y}{1-y}}\right) \qquad \text{This should have been } \color{red}{x=+log_2\left({\frac{y}{1-y}}\right) }\\ \therefore f^{-1}(x)=\color{red}{+} log_2\left({\frac{x}{1-x}}\right)\\ \text{Using the change of base rule this is exactly the same as Heureka's answer.} \)
+69 I tried it by myself an failed...
Then I used the internet and found the answer with steps.
I checked it out in Desmos and it worked. https://www.desmos.com/calculator/qt6xarwxbp
Thank yooooou internet. xD
+118667 Your answer from the internet is no doubt correct, it also matches Heureka's answer I think.
BUT
Your desmos graph [ https://www.desmos.com/calculator/qt6xarwxbp ] is incorrect.
You have graphed it in base 2 NOT in base e.
As you can see, your two graphs are not reflections of one another over the line y=x
That is how I know it is the wrong answer (the answer that you have graphed is wrong I mean)
+26387 Find the inverse of the function
\(\displaystyle f(x)=\frac{2^x}{1+2^x}\)
f(x)=\frac{2^x}{1+2^x}
\(\begin{array}{|rcll|} \hline y &=& \dfrac{2^x}{1+2^x} \quad & | \quad \text{substitute $u =2^x$ } \\\\ y &=& \dfrac{u}{1+u} \\\\ y(1+u) &=& u \\ y+yu &=& u \\ y &=& u-yu \\ y &=& u(1-y) \\\\ u&=& \dfrac{y}{1-y} \quad & | \quad \text{substitute $u =2^x$ } \\\\ 2^x&=& \dfrac{y}{1-y} \quad & | \quad \text{ $\ln{}$ both sides } \\\\ \ln(2^x) &=& \ln\left( \dfrac{y}{1-y} \right) \\\\ x\ln(2) &=& \ln\left( \dfrac{y}{1-y} \right) \\\\ x &=& \dfrac{ \ln\left( \dfrac{y}{1-y} \right) } {\ln(2) } \quad & | \quad x \Leftrightarrow y \\\\ \mathbf{f^{-1}} &\mathbf{=}&\mathbf{ \dfrac{ \ln\left( \dfrac{x}{1-x} \right) } {\ln(2) } } \\ \hline \end{array}\)
