what is the inverse of the function f(x)=2/3x^3-1
the options are
g(x)=^3√6x-6 /2
g(x)=^3√12x+12/2
g(x)=^3√6x+6 /2
g(x)=^3√12x-12/2
THANKS
f(x) =(2/3)x^3 -1
y= (2/3)x^3 - 1
get x by itself...."swap" x and y
add 1 to both sides
y + 1 = (2/3)x^3 mutiply both sides by 3/2
(3/2)y +3/2 = x^3 take the cbe root of both sides
∛ [ (3y + 3) / 2 ] = x
Inside the radical, nutiply numerator, denominator by 4
∛ [ (12y + 12) / 8 ] = x
∛ (12y + 12) / 2 = x "sap x and y
∛ (12x + 12) / 2 = y = the nverse