How do you find the inverse function of #f(x)=x/(x+1)#?
Let
#f^-1(y) = y/(1-y)#
Then:
Hence:
Hence:
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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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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.
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.
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.
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.
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