Question

tìm min , max\(\sqrt{-x^2+2x+4}\)

Answer 1

Đặt \(A=\sqrt{-x^2+2x+4}\)

Ta có \(A\ge0\) theo tính chất căn thức

\(\Rightarrow A_{min}=0\) khi \(-x^2+2x+4=0\Rightarrow x=1\pm\sqrt{5}\)

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

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

Answer 2

\(-x^2+2x-1=-\left(x-1\right)^2\le0\)

\(\Rightarrow-x^2+2x-1+5\le5\)

\(\Rightarrow-x^2+2x+4\le5\)

\(\Rightarrow\sqrt{-x^2+2x+4}\le\sqrt{5}\)

mà \(\sqrt{-x^2+2x+4}\ge2\)

Vậy \(GTLN=\sqrt{5};GTNN=2\)