Question

Сейчас работа тему пропустила

Answer 1

Объяснение:

$1) S=4t+t^2-\frac{t^3}{6}\\ v(t)=S'(t)=(4t+t^2-\frac{t^3}{6})'=4+2t-\frac{1}{6} *3t^2=4+2t-\frac{t^2}{2}\\v(2)=4+2*2-\frac{2^2}{2} =4+4-2=6\\\\a(t)=v'(t)=(4+2t-\frac{t^2}{2})'=2-\frac{1}{2} *2t=2-t\\a(2)=2-2=0\\\\2)x=3t^3+2t+1\\v(t)=x'(t)=(3t^3+2t+1)'=3*3t^2+2=9t^2+2\\v(2)=9*2^2+2=36+2=38\\\\a(t)=v'(t)=(9t^2+2)'=9*2t=18t\\a(2)=18*2=36\\$

$4)tg\alpha =f'(x_{0})\\f(x)=8x-x^4\\f'(x)=8-4x^3\\f'(x_{0})=f'(-2)=8-4*(-2)^3=8+4*8=40\\tg\alpha =40$