Question

Напишите уравнение касательной к графику функции f в точке с абциссой x0 а) f(x) = 3/x, x0 = -1, x0 = 1 б) f(x) = 2x - x^2, x0 = 0, x0 = 2 в) f(x) = x^2 + 1, x0 = 0, x0 = 1 г) f(x) = x^3 - 1, x0 = - 1, x0 = 2

Answer 1

А) 1) f(-$\frac{3}{-1}$1) = -3

2) f ' (x) = $\frac{-3} {x^{2}}$ 

f ' (-1) = $\frac{-3} {-1^{2}}$ = -3

3) y= -3 - 3(x+1) = -3 -3x-3 = -3x -6

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

 2) f ' (x) = $\frac{-3} {x^{2}}$ 

f ' (1) = $\frac{-3} {1^{2}}$ = -3

3) y = 3 - 3(x -1) = 3 - 3x + 3 =  - 3x + 6

б) 1) f(0) = 2*0 - 0 = 0

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

f'(0) = 2 - 2* 0 = 2

3) y = 2x

1) f(2) = 2*2 - 2^{2} = 0

2) f'(x) = (2x)' - $</span(x^{2})'<span$ = 2 - 2x

f'(2) = 2-2*2 = -2

3) y = -2(x-2) = -2x+4