Question

Найти уравнение касательной )

Answer 1

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

1.

$f(x) = {x}^{2} + 2x - 3 \\ x_0 = 1$

$f(1) = 1 + 2 - 3 = 0$

$f'(x) = 2x + 2$

$f'(1) = 4$

$f(x) = 0 + 4(x - 1) = 4x - 4$

- уравнение касательной

2.

$f(x) = {x}^{3} - 3 {x}^{2} + x - 1 \\ x_0 = - 1$

$f( - 1) = - 1 - 3 - 1 - 1 = - 6$

$f'(x) = 3 {x}^{2} - 6x + 1$

$f'( - 1) = 3 + 6 + 1 = 10$

$f(x) = - 6 + 10(x + 1) = - 6 + 10x + 10 = \\ = 10x + 4$

- уравнение касательной

3.

$f(x) = \frac{1}{ {(3x - 8)}^{2} } = {(3x - 8)}^{ - 2} \\ x_0 = 3 \\ \\ f(3) = \frac{1}{1} = 1$

$f'(x) = - 2 {(3x - 8)}^{ - 3} \times 3 = - \frac{6}{ {(3x - 8)}^{3} } \\$

$f'(1) = - \frac{6}{1} = - 6 \\$

$f(x) = 1 - 6(x - 3) = 1 - 6x + 18 = \\ = - 6x + 19$

- уравнение касательной

4.

$f(x) = \frac{ {x}^{2} + 1 }{x} \\ x_0 = 2$

$f(2) = \frac{4 + 1}{2} = 2.5 \\$

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

$f'(2) = \frac{4 - 1}{4} = 0.75 \\$

$f(x) = 2.5 + 0.75(x - 2) = 2.5 + 0.75x - 1.5 = \\ = 0.75x + 1 = \frac{3x}{4} + 1$

- уравнение касательной

5.

$f(x) = 2 \sin( \frac{x}{2} ) \\ x_0 = - \pi$

$f( -\pi ) = 2\sin( - \frac{\pi}{2} ) = - 2 \\$

$f'(x) = \cos( \frac{x}{2} ) \\$

$f'( - \pi) = 0$

$f(x) = - 2$

- уравнение касательной