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]

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]