Question

Запишите уравнение касательно к функции.Задание полностью на фото.​

Answer 1

$f(x) = y(x_0) + y'(x_0)(x - x_0)$

1.

$y = 2 {x}^{2} - 5.5 \\ x_0 = - 0.5$

$y( - 0.5) = \frac{2}{4} - 5.5 = 0.5 - 5.5 = - 5 \\$

$y' = 4x$

$y'( - 0.5) = 4 \times ( - 0.5) = - 2$

$f(x) = - 5 - 2( x + 0.5) = - 5 - 2x - 1 = \\ = - 2x - 6$

2.

$y = 0.2 {x}^{2} - 4 \\ x_0 = 2$

$y(2) = 0.2 \times 4 - 4 = - 3.2$

$y' = 0.4x$

$y'(2) = 0.8$

$f(x) = - 3.2 + 0.8(x - 2) = - 3.2 + 0.8x - 1.6 = \\ = 0.8x - 4.8$

3.

$y = - 3 {x}^{2} - x \\ x_0 = - 2$

$y( - 2) = - 3 \times 4 + 2 = - 10$

$y' = - 6x - 1$

$y'( - 2) = 12 - 1 = 11$

$f(x) = - 10 + 11(x + 2) = - 10 + 11x + 22 = \\ = 11x + 12$

4.

$y = {x}^{2} - \frac{1}{x} \\ x_0 = 3$

$y(3) = 9 - \frac{1}{3} = \frac{26}{3} \\$

$y' = 2x + {x}^{ - 2} = 2x + \frac{1}{ {x}^{2} } \\$

$y'(3) = 6 + \frac{1}{9} = \frac{55}{9} \\$

$f(x) = \frac{26}{3} + \frac{55}{9} (x - 3) = \frac{55x}{9} + \frac{26}{3} - \frac{55}{3} = \\ = \frac{55}{3} x - \frac{29}{3}$