Posted by Anonymous on Thursday, January 10, 2008 at 7:33pm.
Find the inverses of the following functions.
y = 3(x  1)^2, x >= 1
Work:
x = 3(y  1)^2
x = 3(y  1)(y  1)
x = 3(y^2  y  y + 1)
x = 3y^2  6y + 3
And now what do I do!? Please explain and show me how to solve this inverse function!
...
Thank you

Math  Inverse Functions  Damon, Thursday, January 10, 2008 at 7:55pm
x = 3(y  1)^2 would be the inverse but you forgot the second part, y>=1