Question

Нахождение производной в заданной точке 2

Answer 1

Ответ:

$f(x) = \frac{x + 1}{ {x}^{2} } = \frac{1}{x} + \frac{1}{ {x}^{2} } = \\ {x}^{ - 1} + {x}^{ - 2} \\ {f}^{1} (x) = - 1 {x}^{ - 1 - 1} - 2 {x}^{ - 2 - 1} = \\ - \frac{1}{ {x}^{2} } - \frac{2}{ {x}^{3} }$

$- \frac{1}{ {x}^{2} } - \frac{1}{ {x}^{3} } = 0 \\ - \frac{1}{ {x}^{2} } = \frac{1}{ {x}^{3} } \\ {x}^{3} = - {x}^{2} \\ {x}^{3 } + {x}^{2} = 0 \\ {x}^{2} (x + 1) = 0 \\ x = 0 \\ x = - 1$