f(x) = x^3-2
 f(x) = x^4+2
f(a) = f(b)
a = b
The Attempt at a Solution
Assuming that f(a) = f(b), for which some a =/ b
So for  f(x) = x^3-2, let assume a=/b, that f(a) = f(b)
in the end I got, a^3 = b^3
for , I did the same thing, and for f(x) = x^4+2, in the end I got a^4 = b^4
Now I need to simply them down to a = b, so I think I need to take cubic root on both side, and square root on both side, respectively.
Now why is  not a 1-1 function, while  is a 1-1 function?
When I solve for x, for example, in the same of y = x^2, i get sqrt of y = x, so it is not a 1-1 function.
What about x^3-2? I got cub root of y + 2 = x