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How do you find the inverse of #f(x) = x / (x + 8)#?

Answer 1

Sophie Adams

Let #y = f(x)# and solve for #x# to find:

#f^(-1)(y) = (8y)/(1-y)#

#y = f(x) = x/(x+8) = (x+8-8)/(x+8) = 1-8/(x+8)#

Hence (adding #8/(x+8)-y# to both ends):

#8/(x+8) = 1 - y#

Hence (multiplying both sides by #(x+8)/(1-y)#):

#x+8 = 8/(1-y)#

So:

#x = 8/(1-y) - 8 = 8/(1-y) - (8-8y)/(1-y) = (8y)/(1-y)#

So:

#f^(-1)(y) = (8y)/(1-y)#

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Answer 2

Elias Ackerman

To find the inverse of the function ( f(x) = \frac{x}{x + 8} ), follow these steps:

Begin by swapping the roles of ( x ) and ( y ), so the function becomes ( y = \frac{x}{x + 8} ).
Next, solve this equation for ( x ) in terms of ( y ).
Swap ( y ) and ( x ) back to get the inverse function.

So, let's start with ( y = \frac{x}{x + 8} ).

Step 1: Swap ( x ) and ( y ): [ x = \frac{y}{y + 8} ]

Step 2: Solve for ( y ): [ x(y + 8) = y ] [ xy + 8x = y ] [ xy - y = -8x ] [ y(x - 1) = -8x ] [ y = \frac{-8x}{x - 1} ]

Step 3: Swap ( y ) and ( x ) back to get the inverse function: [ f^{-1}(x) = \frac{-8x}{x - 1} ]

So, the inverse function of ( f(x) = \frac{x}{x + 8} ) is ( f^{-1}(x) = \frac{-8x}{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.