“微信扫一扫”进入题库练习及模拟考试
已知函数
当
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 \(\left ( {-\infty ,0} \right )\)