Question

решите неравенство у'> 0 якщо у = x^3+6x^2-15

Answer 1

Ответ:

$x\in ( - \infty ~ ; ~ - 4) \cup ( 0 ~ ; ~ \infty )$

Объяснение:

$\boxed{(x^n)' = n\cdot x^{n-1}}$

$y' > 0 \\\\ (x^3+ 6x^2 - 15)' > 0 \\\\ 3x^2 + 12x > 0 \\\\ 3x(x+4) > 0$

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$x\in ( - \infty ~ ; ~ - 4) \cup ( 0 ~ ; ~ \infty )$