Question

Знайдіть екстремуми функцій
1)f(x)=1/3x^3-x
2)f(x)=x^3/3+3x^2-7x+4

Answer 1

1)

$f(x) = \frac{1}{3} {x}^{3} - x \\ f'(x) = \frac{1}{3} \times 3 {x}^{3 - 1} - 1 = \\ {x}^{2} - 1 = (x - 1)(x + 1) \\ (x - 1)(x + 1) = 0 \\ x _{max} = - 1 \\ x _{min} = 1$

2)

$f(x) = \frac{ {x}^{3} }{3} + 3 {x}^{2} - 7x + 4 \\ f'(x) = \frac{1}{3} \times 3x {}^{3 - 1} + 3 \times 2 x {}^{2 - 1} - 7 = \\ {x}^{2} + 6 {x}^{} - 7 = (x - 1)(x + 7) \\ (x - 1)(x + 7) = 0 \\ x _{max} = - 7 \\ x _{min} = 1$