# if f(x)=2x+1/x+2 findĀ the equation for y=f^-1(x) on which line do the points that are common to the graphs of both functions lie

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### 1 Answer

The inverse function of `f(x) = (2x+1)/(x+2)` has to be determined.

`y = f(x) = (2x+1)/(x+2)`

=> `xy + 2y = 2x + 1`

=> `xy - 2x = 1 - 2y`

=> `x(y - 2) = (1 - 2y)`

=> `x = (1 - 2y)/(y - 2)`

interchange x and y

=> `y = (1 - 2x)/(x - 2)`

The inverse function is `f^-1(x) = (1 - 2x)/(x - 2)`

The graph of `f(x)` and `f^-1(x)` are:

**The equation of the line on which the common points lie is y = x.**