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
6.3x-4-21.6x-0.9=-15.3x-4.9. Подставим x
-51-4.9=-55.9
<span>3/x=-x^2-2x+4 </span>
<span>x<span>∈(-<span>∞;0)U(0;+<span>∞) т.к. на ноль делить нельзя: ( 3/x )</span></span></span></span>