x = 3 y^3 + 2
y^3 = (x-2)/3
so
y = [ (x-2)/3 ]^(1/3) = f-1(x)
so
[ (x-2)/3 ]^(1/3) = 4
[ (x-2)/3 ] = 64
x - 2 = 192
x = 194
Given the function f(x)=3x^3+2, find the value of x so that f^-1(x)=4. Thank you.
I tried to solve for the inverse by susbsituting y in for x so: x=3y^3+2 and then I subtracted 2 so x-2=3y^3 and I am stuck here, do I just cube root the whole thing? If so, do I make it equal to 4?
Thank you.
1 answer
1 answer