f'(x) = (x^2 + 2/x)' = 2x - 2/x^2
2x - 2/x^2 = 0
x - 1/x^2 = 0
(x^3 - 1) / x^2 = 0
x^3 - 1 = 0
x^3 = 1
x = 1
2x - 2/x^2 > 0
x - 1/x^2 > 0
(x^3 - 1) / x^2 > 0
x^3 - 1 = 0 ОДЗ: x^2 = 0
x^3 = 1 x = 0
x = 1
- 0 - 1 +
---○------○---->
x ∈ (1; +∞)
2x - 2/x^2 < 0
x - 1/x^2 < 0
(x^3 - 1) / x^2 < 0
x^3 - 1 = 0 ОДЗ: x^2 = 0
x^3 = 1 x = 0
x = 1
- 0 - 1 +
---○------○---->
x ∈ (-∞; 0) ∪ (0; 1)
Ответ: f'(x) = 0 при x = 1
f'(x) > 0 при x ∈ (1; +∞)
f'(x) < 0 при x ∈ (-∞; 0) ∪ (0; 1)