x∈[-2;3]
x=5y-2
adica y=(x+2)/5
cum x/5 e crescatoare nseamnaca y ia valori intre (-2+2)/5 si (3+2)/5 adica intre 0 si 1/5
y∈[0;1/5]
dar x+2=5y
x²+4x+4+2y²=(x+2)²+2y²=(5y)²+2y²=27y²
a doua expresie de sub radical, separand pe 11 in 9 +2, devine
x²-6x+9 +2y²-4y+2=(x-3)²+2(y-1)²
substitui pe x cu 5y-2
=(5y-2-3)²+2(y-1)²= (5(y-1))²+2(y-1)²=27(y-1)²
deci E= √27y²+√27(y-1)²=√27(|y|+|y-1|)
cum y∈[0;1/5]⇒|y|=y si |y-1|=1-y
atunci E(y)=√27(y+1-y)=√27
=3√3
