f(x) = x² + 1/x - 1
f'(x) = 2x - 1/x² = (2x³ - 1)/x²
Метод интервалов:
0 1/∛2
----о-------·------->
(-) (-) (+)
f(x) ↑, x ∈ (1/∛2; +∞)
f(x) ↓, x ∈ (-∞; 0) ∪ (0; 1/∛2)
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Answers & Comments
f(x) = x² + 1/x - 1
f'(x) = 2x - 1/x² = (2x³ - 1)/x²
Метод интервалов:
0 1/∛2
----о-------·------->
(-) (-) (+)
f(x) ↑, x ∈ (1/∛2; +∞)
f(x) ↓, x ∈ (-∞; 0) ∪ (0; 1/∛2)