To find the critical points of the function f$x$ = 2x^2 - x^8, we first need to find its derivative f'$x$.

f'$x$ = d/dx$2x^2$ - d/dx$x^8$ f'$x$ = 4x - 8x^7

Next, we set f'$x$ = 0 to find the critical points:

4x - 8x^7 = 0
4x = 8x^7
1 = 2x^6
x^6 = 1/2
x = ±$1/2$^$1/6$ x = ±0.629961

So the critical points are at x ≈ ±0.629961.

Therefore, the function f$x$ = 2x^2 - x^8 has critical points at x ≈ ±0.629961.