(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.7(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.71828…是自然对数的底数),曲线y=f(x)在点 (1,f(1))处的切线与x轴平行.(Ⅰ)求k的值;(Ⅱ)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 22:29:47
![(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.7(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.71828…是自然对数的底数),曲线y=f(x)在点 (1,f(1))处的切线与x轴平行.(Ⅰ)求k的值;(Ⅱ)](/uploads/image/z/13877919-63-9.jpg?t=%282012%E5%B1%B1%E4%B8%9C%E6%95%B0%E5%AD%A6%29%28+%2822%29+%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%28lnx%2Bk%29%2Fe%5Ex%28k%E4%B8%BA%E5%B8%B8%E6%95%B0%2Ce%3D2.7%282012%E5%B1%B1%E4%B8%9C%E6%95%B0%E5%AD%A6%29%28+%2822%29+%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%28lnx%2Bk%29%2Fe%5Ex%28k%E4%B8%BA%E5%B8%B8%E6%95%B0%2Ce%3D2.71828%E2%80%A6%E6%98%AF%E8%87%AA%E7%84%B6%E5%AF%B9%E6%95%B0%E7%9A%84%E5%BA%95%E6%95%B0%29%2C%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9+%281%2Cf%281%29%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E4%B8%8Ex%E8%BD%B4%E5%B9%B3%E8%A1%8C.%28%E2%85%A0%29%E6%B1%82k%E7%9A%84%E5%80%BC%EF%BC%9B%28%E2%85%A1%29)
(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.7(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.71828…是自然对数的底数),曲线y=f(x)在点 (1,f(1))处的切线与x轴平行.(Ⅰ)求k的值;(Ⅱ)
(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.7
(2012山东数学)( (22)
已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.71828…是自然对数的底数),曲线y=f(x)在点 (1,f(1))处的切线与x轴平行.
(Ⅰ)求k的值;
(Ⅱ)求的f(x)单调区间;f(x)的求导过程顺便帮忙写一下,我不会╮(╯▽╰)╭
(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.7(2012山东数学)( (22) 已知函数f(x)=(lnx+k)/e^x(k为常数,e=2.71828…是自然对数的底数),曲线y=f(x)在点 (1,f(1))处的切线与x轴平行.(Ⅰ)求k的值;(Ⅱ)
I)函数f(x)=lnx+k ex (k为常数,e=2.71828…是自然对数的底数),∴f′(x)=1 x -lnx-k ex =1-xlnx-kx xex ,x∈(0,+∞),由已知,f′(1)=1-k e =0,∴k=1.
(II)由(I)知,f′(x)=1 x -lnx-1 ex =1-xlnx-x xex ,x∈(0,+∞),设h(x)=1-xlnx-x,x∈(0,+∞),可得h(x)在(0,+∞)上是减函数,又h(1)=0,∴当0<x<1时,h(x)>0,从而f'(x)>0,当x>1时h(x)<0,从而f'(x)<0. 综上可知,f(x)的单调递增区间是(0,1),单调递加区间是(1,+∞)