y=sqrt3*x+sin2x [0;Pi]
Ответы на вопрос
Ответил MiSTiK5267
0
y=f(x0) + f'(x0)(x-x0)
f(x0) и f'(x0)
f(x0) = sin(3*пи/3-2пи/3) = sin(60) = sqrt(3)/2
f'(x0) = 3*cos(3*x0-2пи/3) = 3*cos(3*пи/3-2пи/3) = 3*cos(60) = 3/2
y=sqrt(3)/2 + ( x- пи/3)*3/2
y=(3/2)*x + sqrt(3)/2 - пи/2
f(x0) и f'(x0)
f(x0) = sin(3*пи/3-2пи/3) = sin(60) = sqrt(3)/2
f'(x0) = 3*cos(3*x0-2пи/3) = 3*cos(3*пи/3-2пи/3) = 3*cos(60) = 3/2
y=sqrt(3)/2 + ( x- пи/3)*3/2
y=(3/2)*x + sqrt(3)/2 - пи/2
Новые вопросы