已知函数f(x)=x^2*e^(-x),当曲线y=f(x)的切线I的斜率为负数时,求I在x轴上截距的取值范围 怎么会算...
\u9ad8\u4e2d\u6570\u5b66\u5bfc\u6570\u7684\u5e94\u7528:\u5df2\u77e5\u51fd\u6570fx=x^2*e^(-x),\u5f53\u66f2\u7ebfy=fx\u7684\u5207\u7ebfl\u7684\u659c\u7387\u4e3a\u8d1f\u6570\u65f6\uff0c\u6c42l\u7b54\uff1a
f(x)=x²\u00d7e^(-x)
\u6c42\u5bfc\uff1a
f'(x)=2xe^(-x)-x²\u00d7e^(-x)=x(2-x)e^(-x)<0
\u6240\u4ee5\uff1ax(2-x)<0
\u89e3\u5f97\uff1ax2
\u8bbe\u5207\u70b9\u4e3aP(m\uff0cm²/e^m)\uff0c\u5219\u659c\u7387\u4e3af'(m)=m(2-m)/e^m<0
\u5207\u7ebf\u4e3a\uff1a
y-m²/e^m=m(2-m)(x-m)/e^m
\u4ee4y=0\uff0c\u5f97\u5207\u7ebf\u5728x\u8f74\u4e0a\u7684\u622a\u8dddd=x=(m²-m)/(m-2)
\u6240\u4ee5\uff1a
d=[(m-2)²+3(m-2)+2]/(m-2)
=m-2+2/(m-2)+3
\u5f53m>2\u65f6\uff1ad=m-2+2/(m-2)+3>=2\u221a2+3
\u5f53m2\uff1a
d=m-2+2/(m-2)+3
=-[2-m+2/(2-m)]+3
<-3+3=0
\u7efc\u4e0a\u6240\u8ff0\uff0c\u622a\u8ddd\u7684\u53d6\u503c\u8303\u56f4\u662f(-\u221e\uff0c0) \u222a [3+2\u221a2\uff0c+\u221e)
\u7b54\u9898\u4e0d\u6613\uff0c\u8bb0\u5f97\u91c7\u7eb3\u8c22\u8c22
\uff08\u2160\uff09\u2235f\uff08x\uff09=x 2 e -x \uff0c\u2234f\u2032\uff08x\uff09=2xe -x -x 2 e -x =e -x \uff082x-x 2 \uff09\uff0c\u4ee4f\u2032\uff08x\uff09=0\uff0c\u89e3\u5f97x=0\u6216x=2\uff0c\u4ee4f\u2032\uff08x\uff09\uff1e0\uff0c\u53ef\u89e3\u5f970\uff1cx\uff1c2\uff1b\u4ee4f\u2032\uff08x\uff09\uff1c0\uff0c\u53ef\u89e3\u5f97x\uff1c0\u6216x\uff1e2\uff0c\u6545\u51fd\u6570\u5728\u533a\u95f4\uff08-\u221e\uff0c0\uff09\u4e0e\uff082\uff0c+\u221e\uff09\u4e0a\u662f\u51cf\u51fd\u6570\uff0c\u5728\u533a\u95f4\uff080\uff0c2\uff09\u4e0a\u662f\u589e\u51fd\u6570\uff0e\u2234x=0\u662f\u6781\u5c0f\u503c\u70b9\uff0cx=2\u6781\u5927\u503c\u70b9\uff0c\u53c8f\uff080\uff09=0\uff0cf\uff082\uff09= 4 e 2 \uff0e\u6545f\uff08x\uff09\u7684\u6781\u5c0f\u503c\u548c\u6781\u5927\u503c\u5206\u522b\u4e3a0\uff0c 4 e 2 \uff0e\uff08II\uff09\u8bbe\u5207\u70b9\u4e3a\uff08 x 0 \uff0c x 0 2 e - x 0 \uff09\uff0c\u5219\u5207\u7ebf\u65b9\u7a0b\u4e3ay- x 0 2 e - x 0 = e - x 0 (2 x 0 - x 0 2 ) \uff08x-x 0 \uff09\uff0c\u4ee4y=0\uff0c\u89e3\u5f97x= x 0 2 - x 0 x 0 -2 = (x 0 -2)+ 2 x 0 -2 +3 \uff0c\u56e0\u4e3a\u66f2\u7ebfy=f\uff08x\uff09\u7684\u5207\u7ebfl\u7684\u659c\u7387\u4e3a\u8d1f\u6570\uff0c\u2234 e - x 0 \uff08 2 x 0 - x 20 ) \uff1c0\uff0c\u2234x 0 \uff1c0\u6216x 0 \uff1e2\uff0c\u4ee4 f( x 0 )= x 0 + 2 x 0 -2 +1 \uff0c\u5219 f \u2032 ( x 0 )=1- 2 ( x 0 -2 ) 2 = ( x 0 -2 ) 2 -2 ( x 0 -2 ) 2 \uff0e\u2460\u5f53x 0 \uff1c0\u65f6\uff0c ( x 0 -2 ) 2 -2\uff1e 0\uff0c\u5373f \u2032 \uff08x 0 \uff09\uff1e0\uff0c\u2234f\uff08x 0 \uff09\u5728\uff08-\u221e\uff0c0\uff09\u4e0a\u5355\u8c03\u9012\u589e\uff0c\u2234f\uff08x 0 \uff09\uff1cf\uff080\uff09=0\uff1b\u2461\u5f53x 0 \uff1e2\u65f6\uff0c\u4ee4f \u2032 \uff08x 0 \uff09=0\uff0c\u89e3\u5f97 x 0 =2+ 2 \uff0e\u5f53 x 0 \uff1e2+ 2 \u65f6\uff0cf \u2032 \uff08x 0 \uff09\uff1e0\uff0c\u51fd\u6570f\uff08x 0 \uff09\u5355\u8c03\u9012\u589e\uff1b\u5f53 2\uff1c x 0 \uff1c2+ 2 \u65f6\uff0cf \u2032 \uff08x 0 \uff09\uff1c0\uff0c\u51fd\u6570f\uff08x 0 \uff09\u5355\u8c03\u9012\u51cf\uff0e\u6545\u5f53 x 0 =2+ 2 \u65f6\uff0c\u51fd\u6570f\uff08x 0 \uff09\u53d6\u5f97\u6781\u5c0f\u503c\uff0c\u4e5f\u5373\u6700\u5c0f\u503c\uff0c\u4e14 f(2+ 2 ) = 3+2 2 \uff0e\u7efc\u4e0a\u53ef\u77e5\uff1a\u5207\u7ebfl\u5728x\u8f74\u4e0a\u622a\u8ddd\u7684\u53d6\u503c\u8303\u56f4\u662f\uff08-\u221e\uff0c0\uff09\u222a [2 2 +3\uff0c+\u221e) \uff0e
答：f(x)=x²×e^(-x)
求导：
f'(x)=2xe^(-x)-x²×e^(-x)=x(2-x)e^(-x)<0
所以：x(2-x)<0
解得：x<0或者x>2
设切点为P(m，m²/e^m)，则斜率为f'(m)=m(2-m)/e^m<0
切线为：
y-m²/e^m=m(2-m)(x-m)/e^m
令y=0，得切线在x轴上的截距d=x=(m²-m)/(m-2)
所以：
d=[(m-2)²+3(m-2)+2]/(m-2)
=m-2+2/(m-2)+3
当m>2时：d=m-2+2/(m-2)+3>=2√2+3
当m<0时，m-2<-2，2-m>2：
d=m-2+2/(m-2)+3
=-[2-m+2/(2-m)]+3
<-3+3=0
综上所述，截距的取值范围是(-∞，0) ∪ [3+2√2，+∞)
设切点为（x0，x02e−x0），
则切线方程为y-x02e−x0=e−x0(2x0−x02)（x-x0），
令y=0，解得x=x02−x0x0−2=(x0−2)+2x0−2+3，
因为曲线y=f（x）的切线l的斜率为负数，∴e−x0（2x0−x20)＜0，∴x0＜0或x0＞2，
令f(x0)＝x0+2x0−2+1，
则f′(x0)＝1−2(x0−2)2=(x0−2)2−2(x0−2)2．
①当x0＜0时，(x0−2)2−2＞0，即f′（x0）＞0，∴f（x0）在（-∞，0）上单调递增，∴f（x0）＜f（0）=0；
②当x0＞2时，令f′（x0）=0，解得x0＝2+2．
当x0＞2+2时，f′（x0）＞0，函数f（x0）单调递增；当2＜x0＜2+2时，f′（x0）＜0，函数f（x0）单调递减．
故当x0＝2+2时，函数f（x0）取得极小值，也即最小值，且f(2+2)=3+22．
综上可知：切线l在x轴上截距的取值范围是（-∞，0）∪[22+3，+∞)．
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