F'(x)=0 f(x)=2x^2-x^8

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вопрос

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 =
4x = 8x^
1 = 2x^
x^6 = 1/
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.