<s>[INST] Is this is the right way to do these one-to-one functions, finding their inverse, if not, how to do it? Question 1) 
  $f(x) = 1-x$
My answer (1):   $f(x) = 1-x$, $y = 1 - x$, $y + 1 = x$, $x = y + 1$,  $f$ of inverse $f(y) = y + 1$

Question 2) :  $f(x) = \dfrac{2x}{x-1}$
My answer 2) :  $f(x) = \frac{2x}{x-1} = y$, $\frac{2x}{x-1}$,  $y+1=2x$, $\frac{y+1}{2} = x$, $x= \frac{y+1}{2}$, $f$ of inverse $f(y) = \frac{y+1}{2}$

Question 3):  $f(x) = \sqrt{5} - x$ 
My answer 3) : $f(x) = \sqrt{5} - x = y = \sqrt{5} - x = y + 5 = x = x = y + 5$, f of inverse $f(y) = y + 5$

Question 4):  $f(x) = x^3$
My answer 4): $f(x) = x^3$,  $y=x^3$, $y^3 = x$, $x=y^3$, f of inverse $f(y)  = y^3$
 [/INST] Question 1. I believe the method is right, but there is a small mistake. 
\begin{align*}
y &= 1-x \\
y-1 &= -x \\
1-y &= x
\end{align*}

Question 2. 
\begin{align*}
y &= \frac{2x}{x+1} \\
\frac{y}{2} &= \frac{x}{x+1} \\
\frac{y}{2} &= 1-\frac{1}{x+1} \\
\frac{y}{2}-1 &= -\frac{1}{x+1} \\
1-\frac{y}{2} &= \frac{1}{x+1} \\
\frac{1}{1-\frac{y}{2}} &= x+1 \\
\frac{1}{1-\frac{y}{2}}-1 &= x \\
\end{align*}

Question 3.
This is similiar to question 1, can you do this yourself?
\begin{align*}
y &= \sqrt{5}-x \\
y-\sqrt{5} &= -x \\
\sqrt{5}-y &= x
\end{align*}

Question 4.
\begin{align*}
y &= x^3 \\
\sqrt[3]{y} &= x
\end{align*}
</s>