Can't put expression in terms of $y:\ x = \frac{2y-1}{y+1}.$
Disclaimer: This is not a student posting his homework assignment.
I am an adult learning Calculus. I think this is a great forum,
Ok, you know the formula for the derivative of an inverse:
$f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$
Using this, how do you answer this one?
Find $(f^{-1})'(x)$ if $f(x)=\frac{2x-1}{x+1}$
First, using Chain rule, I determined that $f'(x) = \frac{3}{(x+1)^2}$
But, I was having trouble getting the inverse of f(x).
$y=\frac{2x-1}{x+1}$
Inverse is: $x=\frac{2y-1}{y+1}$
But, I can not put this in terms of y! What do I do?
$\endgroup$1 Answer
$\begingroup$Hint: It seems you have swapped the $x$ and $y$. Now multiply by $y + 1$ to obtain $$xy + x = 2y-1.$$ Given that you are trying to isolate $y$, put all the terms with a $y$ in it on the left, and all other terms on the right. Then take a common factor of $y$ out of the expression on the left hand side; the terms that remain do not contain any $y$ terms. Now there is only one other step to obtain an expression for $y$.
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