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高中数学选择性必修 第二册(381题)


已知函数 \(h\left ( {x} \right )=x{e}^{x}-mx,g\left ( {x} \right )=\text{ln}x+x+1\).


\( m=1\) 时,求函数\( h\left(x\right)\)的单调区间



知识点:第五章 一元函数的导数及其应用


参考答案:当 \( m=1\) 时 \(h\left ( {x} \right )=x{e}^{x}-x{,h}^{'}\left ( {x} \right )=\left ( {x+1} \right ){e}^{x}-1\)

设 \( \mu \left(x\right)={h}^{\text{'}}\left(x\right)=\left(x+1\right){e}^{x}-1\),则 \( {\mu }^{\text{'}}\left(x\right)=\left(x+2\right){e}^{x}\ge 0\Rightarrow x\ge -2\)

\( \begin{array}{c}{\mu }^{\text{'}}\left(x\right)=\left(x+2\right){e}^{x}<0\Rightarrow x<-2\end{array}\) \( \therefore \mu \left(x\right)\) 即 \( {h}^{\text{'}}\left(x\right)\) 在 \(\left ( {-\infty ,-2} \right )\) 递减,在 \(\left . {[-2,+\infty } \right )\) 递增,

当 \(x\in \left ( {-\infty ,-2} \right ),{h}^{'}\left ( {x} \right )=\left ( {x+1} \right )\cdot {e}^{x}-1<0\) ,当 \(x\in [\left . {-2,0} \right ),{h}^{'}\left ( {x} \right )<{h}^{'}\left ( {0} \right )=0\)

而当 \(x\in \left . {[0,+\infty } \right ),{h}^{'}\left ( {x} \right )\ge {h}^{'}\left ( {0} \right )=0\) 所以当 \(x\in \left ( {-\infty ,0} \right ),{h}^{'}\left ( {x} \right )<0,h\left ( {x} \right )\) 递减;

\(x\in \left . {[0,+\infty } \right ),{h}^{'}\left ( {x} \right )\ge 0,h\left ( {x} \right )\) 递增.故函数增区间为 \(\left . {[0,+\infty } \right )\) ,减区间为 \(\left ( {-\infty ,0} \right )\)

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