f(x)+1=1(x+1)+1√3−2x−x2≤√8 (BĐT Bunhia)Min :$f(x)=-\sqrt{8}-1\Leftrightarrow x=-\sqrt{2}-1$Max :$f(x)=\sqrt{8}-1\Leftrightarrow x=\sqrt{2}-1$
Điều kiện: −3≤x≤1f(x)+1=1(x+1)+1√3−2x−x2≤√8 (BĐT Bunhia)M
ax :
f(x)=√8−1⇔x=√2−1M
in: $f(x)=-
3\Leftrightarrow x=-
3$