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