Question

пожалуйста, помогите решить

Answer 1

Ответ:

$tg \alpha = k = f'(x0)$

1.

а)

$f'(x) = 2 \times 3 {x}^{2} + 8 \times 2x + 1 = 6 {x}^{2} + 16x + 1$

$f'( - 3) = 6 \times 9 - 16 \times 3 + 1 = \\ = 54 - 48 + 1 = 7$

б)

$f'(x) = - \sin(x) \\ f'( \frac{\pi}{6} ) = - \sin( \frac{\pi}{6} ) = - \frac{1}{2} = - 0.5$

в)

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

$f'(2) = \frac{1}{ \sqrt{2} } = \frac{ \sqrt{2} }{2}$

г)

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

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

2.

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

а)

$f(x) = 6 {x}^{3} - 2 {x}^{2} + 5x \\ x_0 = 1$

$f(1) = 6 - 2 + 5 = 9$

$f'(x) = 18 {x}^{2} - 4x + 5$

$f'(1) = 18 - 4 + 5 = 19$

$f(x) = 9 + 19(x - 1) = 9 + 19x - 19 = \\ = 19x - 10$

- уравнение касательной

б)

$f(x) = \frac{3}{x} \\ x_0 = - 3$

$f( - 3) = \frac{3}{ - 3} = - 1$

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

$f'( - 3) = - \frac{3}{9} = - \frac{1}{3}$

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

- уравнение касательной

в)

$f(x) = \sin(x) \\ x_0 = \frac{\pi}{6}$

$f( \frac{\pi}{6} ) = \sin( \frac{\pi}{6} ) = \frac{1}{2}$

$f'(x) = \cos(x)$

$f'( \frac{\pi}{6} ) = \cos( \frac{\pi}{6} ) = \frac{ \sqrt{3} }{2}$

$f(x) = \frac{1}{2} + \frac{ \sqrt{3} }{2} (x - \frac{\pi}{6} ) = \\ = \frac{1}{2} + \frac{ \sqrt{3} }{2} x - \frac{\pi \sqrt{3} }{12} = \\ = \frac{ \sqrt{3} }{2} x + \frac{6 - \pi \sqrt{3} }{12}$

- уравнение касательной