# How do we find the inverse of the function #y=e^(2x)/(7+e^(2x))#?

Inverse function of

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To find the inverse of the function ( y = \frac{e^{2x}}{7 + e^{2x}} ), follow these steps:

- Replace ( y ) with ( x ) and ( x ) with ( y ).
- Solve the resulting equation for ( y ).
- Swap ( x ) and ( y ) to express the inverse function.

Let's proceed with the calculations:

Original function: ( y = \frac{e^{2x}}{7 + e^{2x}} )

- Replace ( y ) with ( x ) and ( x ) with ( y ):

[ x = \frac{e^{2y}}{7 + e^{2y}} ]

- Solve for ( y ):

[ x(7 + e^{2y}) = e^{2y} ] [ 7x + xe^{2y} = e^{2y} ] [ xe^{2y} - e^{2y} = -7x ] [ e^{2y}(x - 1) = -7x ] [ e^{2y} = \frac{-7x}{x - 1} ] [ 2y = \ln\left(\frac{-7x}{x - 1}\right) ] [ y = \frac{1}{2}\ln\left(\frac{-7x}{x - 1}\right) ]

- Swap ( x ) and ( y ):

[ \boxed{f^{-1}(x) = \frac{1}{2}\ln\left(\frac{-7x}{x - 1}\right)} ]

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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.

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