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 = 04x = 8x^71 = 2x^6x^6 = 1/2x = ±(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.
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.