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What is the inverse function of f(x)=x over x+1 . sorry , i dont know how to make the fraction sign .
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\[f(z)=\frac{x}{x+1}\]
set \[x=\frac{y}{y+1}\] and solve the equation for \(y\) that will be your inverse
any idea how to do that?
not a clue
first get rid of the fraction by multiplying both sides by \(y+1\) to get \[x(y+1)=y\]
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then multiply out on the left to get \[xy+x=y\]
put all the \(y\) terms on one side by subtracting \(xy\) and get \[x=y-xy\]
factor out the \(y\) \[x=y(1-x)\]
divide both sides by \(1-x\) and finish with \[y=\frac{x}{1-x}\] and that is your inverse
thank you ;)
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yw hope it is clear, all steps are there
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