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
А) 1) f(-1) = -3
2) f ' (x) =
f ' (-1) = = -3
3) y= -3 - 3(x+1) = -3 -3x-3 = -3x -6
1) f(-1) = 3
f ' (1) = = -3
3) y = 3 - 3(x -1) = 3 - 3x + 3 = - 3x + 6
б) 1) f(0) = 2*0 - 0 = 0
2) f'(x) = (2x)' - = 2 - 2x
f'(0) = 2 - 2* 0 = 2
3) y = 2x
1) f(2) = 2*2 - 2^{2} = 0
2) f'(x) = (2x)' - (x^{2})'" alt="(x^{2})'" align="absmiddle" class="latex-formula"> = 2 - 2x
f'(2) = 2-2*2 = -2
3) y = -2(x-2) = -2x+4