y=√1-x°2 tính cực đại và cực tiểu

\[\begin{array}{l} y = \sqrt {1 - {x^2}} \\ TXD:\,\,\,D = \left[ { - 1;\,\,1} \right]\\ \Rightarrow y' = \frac{{ - x}}{{2\sqrt {1 - {x^2}} }}\\ \Rightarrow - x = 0\\ \Leftrightarrow x = 0\\ Bang\,\,\,xet\,\,\,dau:\\ - 1\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ y\left( 0 \right) = 1;\,\,\,y\left( { - 1} \right) = y\left( 1 \right) = 0\\ \Rightarrow x = 0\,\,\,la\,\,\,diem\,\,\,cuc\,\,dai\,\,cua\,\,ham\,\,so. \end{array}\]