Question

Tìm GTLN, GTNN của B=\(\sqrt{-x^2+2x+4}\)

Answer 1

\(ĐK:1-\sqrt{5}\le x\le1+\sqrt{5}\)

\(B=\sqrt{-x^2+2x+4}=\sqrt{-\left(x^2-2x-4\right)}=\sqrt{-\left(x^2-2x+1\right)+5}=\sqrt{-\left(x-1\right)^2+5}\le\sqrt{5}\)

\(\Rightarrow Max_B=\sqrt{5}\Leftrightarrow x=1\)