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How do you find the inverse of #f(x) = (2x-1)/(x-1)#?

Answer 1

Serenity Johnson

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

#f^(-1)(y) = (y-1)/(y-2)#

#y = f(x) = (2x-1)/(x-1) = (2x-2+1)/(x-1) = (2(x-1)+1)/(x-1)#

#= 2+1/(x-1)#

Subtract #2# from both ends to get:

#y-2 = 1/(x-1)#

Hence:

#x-1 = 1/(y-2)#

Add #1# to both sides to get:

#x = 1/(y-2)+1 = (1+(y-2))/(y-2) = (y-1)/(y-2)#

So #f^(-1)(y) = (y-1)/(y-2)#

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Answer 2

Ariana Alvarez

To find the inverse of ( f(x) = \frac{2x - 1}{x - 1} ), follow these steps:

Replace ( f(x) ) with ( y ): ( y = \frac{2x - 1}{x - 1} ).
Swap ( x ) and ( y ): ( x = \frac{2y - 1}{y - 1} ).
Solve this equation for ( y ) to find the inverse.

[ x = \frac{2y - 1}{y - 1} ] [ x(y - 1) = 2y - 1 ] [ xy - x = 2y - 1 ] [ xy - 2y = x - 1 ] [ y(x - 2) = x - 1 ] [ y = \frac{x - 1}{x - 2} ]

Therefore, the inverse function of ( f(x) = \frac{2x - 1}{x - 1} ) is ( f^{-1}(x) = \frac{x - 1}{x - 2} ).

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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.