i got y= [(x-3)^1/3]/2
but I'm pretty sure i've messed up the order of operations and im not sure if its supposed to switch to x =...
2007-09-14 12:40:40 · 5 answers · asked by Ar J 2 in Science & Mathematics ➔ Mathematics
y = 2x^3 + 3
subtract 3
y - 3 = 2x^3
divide by 2
(y - 3)/2 = x^3
x = [(y - 3)/2]^1/3
f^-1(x) = [(x - 3)/2]^(1/3)
2007-09-14 12:52:34 · answer #1 · answered by mohanrao d 7 · 0⤊ 0⤋
f inverse = { (x-3)/2}^(1/3)
Steps: subtract 3, divide by 2, cube root last.
2007-09-14 12:45:59 · answer #2 · answered by jenh42002 7 · 0⤊ 0⤋
y=2x^3+3
or y-3=2x^3
or x^3={(y-3)/2}
or x={(y-3)/2}^1/3
swithing to x,we get
or f(inverse)={(x-3)/2}^1/3 ans
2007-09-14 13:31:38 · answer #3 · answered by MAHAANIM07 4 · 0⤊ 0⤋
y = [(x-3)/2]^(1/3)
You must divide 2 first and then take cubic root.
2007-09-14 12:46:29 · answer #4 · answered by sahsjing 7 · 0⤊ 0⤋
y=cubrt((x-3)/2)
2007-09-14 12:51:42 · answer #5 · answered by bassplayajermz 2 · 0⤊ 0⤋