1.Найдите тангенс угла наклона к касательной к графику функции у=f(x) в точке Xo: a) f(x)=4cosx+x, Xo=П/6 б)f(x)=3x^2-12x+5

1.

a) f(x) = 4cosx + x, Xo = П/6.

tg a = f\'(x0).

f\'(x) = -4sinx + 1,

f\'(П/6) = -4 sinaП/6 + 1 = -4 *0.5 + 1 = -2 + 1 = -1.

tg a = -1.

б)f(x) = 3x^2 - 12x + 5, Xo = -12,

f\'(x) = 6x - 12,

f\'(12) = 6 * 12 - 12 = 60.

tg a = 60.

2. y = f\'(x0) * (x - x0) + f(x0).

a) f(x) = 2x^3 - x, Xo = -2.

f(-2) = 2 * (-8) + 2 = -14.

f\'(x) = 6x^2 - 1,

f\'(-2) = 23.

y = 23 * (x + 2) -14 = 23x + 46 - 14 = 23x + 32,

y = 23x + 32.

б) f(x) = ln(3x - 2), Xo = 1.

f(1) = ln(3 - 2) = 0.

f\'(x) = 1 / (3 * (3x - 2)),

f\'(1) = 1 / (3 * (3 - 2)) = 1/3.

y = 1/3 * (x - 1),

y = 1/3 * x - 1/3.