you need to interchange the domain and range variables and then solve that again for the range variable...ex y = 5x + 2---> x = 5 y + 2---> [x-2] / 5 = y for the inverse....your answers are contained in { ( [ x + 10] / 5 ) ^ (1/3) , [ x/3]^(2/5), [ 7x + 3] / 4, [x-2] / 2 ,
in case you mean y = 2 - (x/3) + x, then (2-y) = (x/3) - x 2-y = 4x/3 x=(3/4)(2-y) in case you mean f(x) = y = (2-x)/(x+3), then y(x+3) = 2-x xy + 3y = 2 - x xy + x = 2 - 3y x(y+one million) = 2 - 3y x = (2-3y)/(y+one million)
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you need to interchange the domain and range variables and then solve that again for the range variable...ex y = 5x + 2---> x = 5 y + 2---> [x-2] / 5 = y for the inverse....your answers are contained in { ( [ x + 10] / 5 ) ^ (1/3) , [ x/3]^(2/5), [ 7x + 3] / 4, [x-2] / 2 ,
[ 3 x] ^5, [2/x] + 10 , sin [x -5]}
in case you mean y = 2 - (x/3) + x, then (2-y) = (x/3) - x 2-y = 4x/3 x=(3/4)(2-y) in case you mean f(x) = y = (2-x)/(x+3), then y(x+3) = 2-x xy + 3y = 2 - x xy + x = 2 - 3y x(y+one million) = 2 - 3y x = (2-3y)/(y+one million)