Ответ:
[1,5; 4]
Объяснение:
f(x)=|x²-3x+2|+|x²+x|, x∈[-1/2; 3/2]
x²-3x+2=x²-2x-x+2=x(x-2)-(x-2)=(x-2)(x-1)
f(x)=|x²-3x+2|+|x²+x|=f(x)=|(x-2)(x-1)|+|x(x+1)|
x∈[-1/2; 0]⇒(x-2)(x-1)>0; x(x+1)<0⇒f(x)=x²-3x+2-x²-x=-4x+2⇒E₁(f)=[2;4]</p>
x∈[0; 1]⇒(x-2)(x-1)≥0; x(x+1)≥0⇒f(x)=x²-3x+2+x²+x=2x²-2x+2=
=2(x-0,5)²+1,5⇒E₂(f)=[1,5; 2]
x∈[1; 1,5]⇒(x-2)(x-1)<0; x(x+1)>0⇒f(x)=-x²+3x-2+x²+x=4x-2⇒E₃(f)=[2;4]
E(f)=E₁∪E₂∪E₃=[1,5; 4]