Question

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4. Найти f′(x0), если f(x) = (1 − x)^2(x + 2), x0 = −2.

5. Найти f′(x0), если f(x) =x^3/(x−2)^2, x0 = 6.

6. Найти f′(x0), если f(x) = sin^2πx ∙ (1 −x/2) , x0 = 6.

Answer 1

Ответ:

$4)f'(x) = 2(1 - x) \times ( - 1) \times (x + 2) + {(1 - x)}^{2} = \\ = (1 - x) \times ( - 2(x + 2) + 1 - x) = \\ = (1 - x) \times ( - 2x - 4 + 1 - x) = (1 - x) \times ( - 3x - 3)$

$f'( - 2) = (1 + 2) \times (6 - 3) = 3 \times 3 = 9$

$5)f'(x) = 3 {x}^{2} \times {(x - 2)}^{2} + 2(x - 2) {x}^{3} = \\ = {x}^{2} (x - 2) \times (3(x - 2) + 2x) = \\ = {x}^{2} (x - 2)(3x - 6 + 2x) = \\ = {x}^{2} (x - 2)(5x - 6)$

$f'(6) = 36 \times 4 \times (30 - 6) = 36 \times 4 \times 24 = 576$

$6)f'(x) = 2 \sin(\pi \: x) \times \cos(\pi \: x) \times (1 - \frac{x}{2} ) + ( - \frac{1}{2} ) { \sin }^{2} (\pi \: x) = \\ = ( 1- \frac{x}{2} ) \sin(2\pi \: x) - \frac{1}{2} { \sin }^{2} (\pi \: x)$

$f'(6) = (1 - 3) \sin(8\pi) - \frac{1}{2} { \sin }^{2} (6\pi) = 0 - 0 = 0$