Задача 79941 ...
Условие
Решение
[m]\large f(-1) = \frac{(-1)^2 - 5(-1) - 1}{(-1)^3} = \frac{1 + 5 - 1}{-1} = \frac{5}{-1} = -5[/m]
Находим производную от дроби:
[m]\large f'(t) = \frac{(t^2 - 5t - 1)' t^3 - (t^2 - 5t - 1)(t^3)'}{(t^3)^2} = \frac{(2t - 5) t^3 - (t^2 - 5t - 1) \cdot 3t^2}{t^6} = [/m]
[m]\large = \frac{(2t - 5) t - 3(t^2 - 5t - 1)}{t^4} = \frac{2t^2 - 5t - 3t^2 + 15t + 3}{t^4} = \frac{-t^2 + 10t + 3}{t^4}[/m]
[m]\large f'(-1) = \frac{-(-1)^2 + 10(-1) + 3}{(-1)^4} = \frac{-1 - 10 + 3}{1} = -8[/m]
[m]\large f'(2) = \frac{-2^2 + 10 \cdot 2 + 3}{2^4} = \frac{-4 + 20 + 3}{16} = \frac{19}{16}[/m]
[m]\large f'(\frac{1}{a}) = \frac{-(1/a)^2 + 10 \cdot 1/a + 3}{(1/a)^4} = a^4 \cdot (-\frac{1}{a^2} + \frac{10}{a} + 3) = -a^2 + 10a^3 + 3a^4[/m]