Học lớp hướng dẫn giải
TXĐ: \(D = \left[ { - \sqrt 5 ;\sqrt 5 } \right]\)
\(\begin{array}{l} y' = 2 + \frac{{ - 2x}}{{2\sqrt {5 - {x^2}} }}\\ y' = 0 \Leftrightarrow 2 - \frac{x}{{\sqrt {5 - {x^2}} }} = 0 \Leftrightarrow \left\{ \begin{array}{l} x = 2\sqrt {5 - {x^2}} \\ \sqrt {5 - {x^2}} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 0 \le x < \sqrt 5 \\ {x^2} = 4\left( {5 - {x^2}} \right) \end{array} \right. \Leftrightarrow x = 2 \end{array}\)
Ta có:
\(\begin{array}{l} y( - \sqrt 5 ) = - 2\sqrt 5 ;\,\,\,\,y\left( 2 \right) = 5\\ y\left( {\sqrt 5 } \right) = 2\sqrt 5 \\ \Rightarrow \mathop {Max}\limits_{x \in D} y = 5 \end{array}\)
