Question

Tìm Min,Max

B=\(\sqrt{-x^2+2x+4}\)

Answer 1

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

\(\Rightarrow B_{max}=\sqrt{5}\) khi \(x=1\)

ta có : \(B=\sqrt{-x^2+2x+4}\ge0\)

\(\Rightarrow B_{min}=0\) khi \(-x^2+2x+4=0\Leftrightarrow\left[{}\begin{matrix}1+\sqrt{5}\\1-\sqrt{5}\end{matrix}\right.\)

vậy .............................................................................................................