- Min:
\(A=x\sqrt{4-x^2}\)
ĐK: \(-2\le0\le2\)
\(\Rightarrow2A=2x\sqrt{4-x^2}\)
\(=x^2+2x\sqrt{4-x^2}+4-x^2-4\)
\(=\left(x+\sqrt{4-x^2}\right)^2-4\ge-4\)
\(\Rightarrow A\ge-2\)
Dấu \("="\) xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{4-x^2}\\x\sqrt{4-x^2}=-2\end{matrix}\right.\)
\(\Rightarrow x^2=4-x^2\)
\(\Rightarrow2x^2=4\)
\(\Rightarrow x^2=2\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\left(ktm\right)\\x=-\sqrt{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(A_{min}=-2\) \(\Leftrightarrow x=-\sqrt{2}\)
- Max:
Ta có: \(\sqrt{x^2\left(4-x^2\right)}\le\frac{x^2+4-x^2}{2}=2\)
Dấu \("="\) xảy ra \(\Leftrightarrow x^2=4-x^2\)
\(\Leftrightarrow2x^2=4\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\left(tm\right)\\x=-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)
Vậy \(A_{max}=2\) \(\Leftrightarrow x=\sqrt{2}\)
