Ta có: \(x \in \left[ { - 3;3} \right] \Rightarrow 0 \le {x^2} \le 9 \Rightarrow 0 \le 9 - {x^2} \le 9 \Rightarrow 0 \le \sqrt {9 - {x^2}} \le 3\).
Vậy \(\left\{ \begin{array}{l}\mathop {\max }\limits_{\left[ { - 3;3} \right]} f\left( x \right) = 3 \Leftrightarrow x = 0\\\mathop {\min }\limits_{\left[ { - 3;3} \right]} f\left( x \right) = 0 \Leftrightarrow x = \pm 3\end{array} \right.\).
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