Question

Запишите уравнения касательной к функции y = f(x) при x = x0

Answer 1

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

- формула для поиска

1.

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

$y( - 0.5) = 2 \times 0.25 - 5.5 = 0.5 - 5.5 = - 5 \\$

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

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

- уравнение касательной

2.

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

$y(2) = 0.2 \times 4 - 4 = 0.8 - 4 = - 3.2$

$y '= 0.4x \\ y'(2) = 0.8$

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

- уравнение касательной

3.

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

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

$y '= - 6x - 1 \\ yx( - 2) = 12 - 1 = 11$

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

- уравнение касательной

4.

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

$y(3) = 9 - \frac{1}{3} = \frac{11}{3} \\$

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

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

$f(x) = \frac{11}{3} + \frac{55}{9}(x + 3) = \frac{11}{3} + \frac{55}{9}x + \frac{55}{3} \\ f(x) = \frac{55}{9} x + \frac{66}{3} \\ f(x) = \frac{55}{9} x + 22$

- уравнение касательной