Question

Составьте уравнение касательной к графику функции fв точке с абсциссой x0 если: f(x)=x^3-5x,x0=2
2) f(x) =1\x+1, x0=-2
3) f(x) = корень x-3,x0=4
4)f(x)=cos2x, x0=п\8

Answer 1

Ответ:

В объяснении

Объяснение:

Формула уравнения касательной:

y = f(x0) + f ` (x0) × (x-x0)

1) f(x) = x^3 - 5x

x0 = 2

f ` (x) = (x^3 - 5x) ` = 3x^2 - 5

f(2) = 2^3 - 5 × 2 = 8 - 10 = -2

f ` (2) = 3 × 2^2 - 5 = 12 - 5 = 7

y = (-2) + 7 × (x - 2) = (-2) + 7x - 14 = 7x - 16

2) f(x) = 1/x + 1

x0 = -2

f ` (x) = (1/x + 1) ` = (x^-1 + 1) ` = (-x^-2) =

$- \frac{1}{ {x}^{2} }$

f (-2) = 1/(-2) + 1 = (-0.5) + 1 = 0.5

f ` (-2) =

$- \frac{1}{ {( - 2)}^{2} } = - \frac{1 }{4} = - 0.25$

y = 0.5 - 0.25 × (x+2) = 0.5 - 0.25x + 0.5 = 1 - 0.25x

3) f(x) =

$\sqrt{x - 3}$

*Примечание , все под корнем.

x0 = 4

f ` (x) = (√x-3) ` =

$\frac{1}{2 \sqrt{x - 3} } \times 1 = \frac{1}{2 \sqrt{x - 3} }$

f (4) = √4-3 = √1 = 1

f ` (4) =

$\frac{1}{2 \sqrt{4 - 3} } = \frac{1}{2 \sqrt{1} } = \frac{1}{2} = 0.5$

y = 1 + 0.5 × (x - 4) = 1 + 0.5x - 2 = 0.5x - 1

4) f(x) = cos2x

x0 = π/8

f ` (x) = (cos2x) ` = (-sin2x) × 2 = -2sin2x

f (π/8) = cos2 × π/8 = cos(π/4) =

$\frac{ \sqrt{2} }{2}$

f ` (π/8) = -2sin2 × π/8 = -2 × √2/2 = - √2

$y = \frac{ \sqrt{2} }{2} - \sqrt{2} \times (x - \frac{\pi}{8} ) = \frac{ \sqrt{2} }{2} - \sqrt{2}x + \frac{ \sqrt{2} \pi}{8}$