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How do you find the inverse function of #f(x)=x/(x+1)#?

Answer 1

Madelyn Anderson

Let #y = f(x)# and rearrange to find an expression for #x# in terms of #y#, hence:

#f^-1(y) = y/(1-y)#

Let #y = f(x) = x/(x+1) = (x+1-1)/(x+1) = 1-1/(x+1)#

Then:

#1-y = 1/(x+1)#

Hence:

#1/(1-y) = x+1#

Hence:

#x = 1/(1-y)-1 = (1-(1-y))/(1-y) = y/(1-y)#

So #f^-1(y) = y/(1-y)#

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Answer 2

Cora Alexander

To find the inverse function of ( f(x) = \frac{x}{x+1} ), follow these steps:

Replace ( f(x) ) with ( y ).
Swap ( x ) and ( y ).
Solve the resulting equation for ( y ).
Replace ( y ) with ( f^{-1}(x) ) to express the inverse function.

Following these steps:

[ y = \frac{x}{x+1} ] [ x = \frac{y}{y+1} ] [ x(y+1) = y ] [ xy + x = y ] [ xy - y = -x ] [ y(x - 1) = -x ] [ y = \frac{-x}{x - 1} ]

So, the inverse function of ( f(x) = \frac{x}{x+1} ) is ( f^{-1}(x) = \frac{-x}{x - 1} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.