# If g(x)=x/(3x+2) then find the inverse of g and give the domain and range of the inverse of g using interval notation?

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You need to write down the original function `y = x/(3x+2)` and then you should solve for x the equation `y = x/(3x+2)` such that:

`3xy + 2y = x =gt x - 3xy = 2y `

You need to factor out x such that:

`x(1 - 3y) = 2y =gt x = (2y)/(1-3y)`

You need to switch x and y in equation `x = (2y)/(1-3y)` such that:

`y = (2x)/(1-3x)`

The domain of the function consists of all values of x that makes the function possible. Notice that the function is rational, hence the denominator can have any real value, except 0.

You need to find what value of x cancels denominator such that:

`1 - 3x = 0 =gt -3x = -1 =gtx = 1/3`

**Hence, the domain of inverse function `g^(-1)(x) = (2x)/(1-3x)` is `(-oo,1/3)U(1/3,oo)` and the range is the real set R.**

Let g(x)=y

y=x/(3x+2)

= 1/3 - 2/(3(3x+2))

=> y-1/3=2/(9x+6)

=> 9x+6=2/(-y+1/3)

= 6/(1-3y)

=> 9x=6/(1-3y)-6

=18y/(1-3y)

=> x= 2y/(1-3y)

Therefore, the inverse of function g(x) would be y=2x/(1-3x)

since the denominator cannot be zero, 1-3x cannot be zero.

so, x cannot be 1/3

Therefore, the domain of the function is all real number except 1/3, and the range would be all real number except -2/3

(because from the original g(x), 3y+2 cannot be zero and y cannot be -2/3)