v(t)=t²-4t+3
a(t)=2t-4
2t-4=0
2t=4
t=2
Ответ: 3.2
=а²-4-а²-1=-4-1=-(4+1)=-5
F(x)=4x+2x²-x³ [0;3]
f`(x)=4-4x-3x²=0
3x²+4x-4=0 D=64
x=2/3 x=-2 ∉ [0;3]
f(0)=4*0+2*0²-0³=0
f(2/3)=4*(2/3)+2*(2/3)²-(2/3)³=8/3+8/9-8/27=(8*9+8*3-8)/27=
=(72+24-8)/27=88/27=3⁷/₂₇=max
f(3)=4*3+2*3²-3³=12+18-27=3
2x+1=3x-4,2x-3x=-4-1, -x=-5, x=5