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高中数学选择性必修 第二册(381题)


已知函数 \(f\left( x \right) = {x^3} - 3ax + 4(a > 0)\) ,定义域为 \(\left[ { - 2, + \infty } \right)\)


若函数\(f\left ( {x} \right )\)在区间 \(\left[ { - 2,1} \right]\) 上的最大值为20,求实数a的值.




知识点:第五章 一元函数的导数及其应用


参考答案:\(f'\left ( {x} \right )=3\left ( {{{x}^{2}}-a} \right )\)

①若 \(0 < a < 1\) ,\(x \in \left( { - 2, - \sqrt a } \right)\), \(f'\left ( {x} \right )>0\) , \(f\left( x \right)\) 单调递增

\(x \in \left( { - \sqrt a ,\sqrt a } \right)\),\(f\prime \left( x \right) < 0\),\(f\left( x \right)\) 单调递减,\(x \in \left( {\sqrt a ,1} \right)\),\(f'\left ( {x} \right )>0\), \(f\left( x \right)\) 单调递增

若 \(f{\left( x \right)_{\max }} = f\left( { - \sqrt a } \right) = 2a\sqrt a + 4 = 20\) ,得 \(a = 4\) (舍去)

若 \(f{\left( x \right)_{\max }} = f\left( 1 \right) = 5 - 3a = 20\) ,得 \(a = - 5\) (舍去)

②若 \(1 \leqslant a < 4\), \(x \in \left( { - 2, - \sqrt a } \right)\) ,\(f'\left ( {x} \right )>0\), \(f\left( x \right)\) 单调递增

\(x \in \left( { - \sqrt a ,1} \right)\),\(f\prime \left( x \right) < 0\),\(f\left( x \right)\) 单调递减

若 \(f{\left( x \right)_{\max }} = f\left( { - \sqrt a } \right) = 2a\sqrt a + 4 = 20\) ,得 \(a = 4\) (舍去)

③若 \(a \geqslant 4\) ,\(x \in \left( { - 2,1} \right)\),\(f'\left ( {x} \right )<0\) ,\(f\left( x \right)\) 单调递减

若 \(f{\left( x \right)_{\max }} = f\left( { - 2} \right) = 6a - 4 = 20\) ,得 \(a = 4\) (满足)

综上所述: \(a = 4\)

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