设函数f(x)=ax2+bx+k(k>0)在x=0处取得极值,且曲线y=f(x)在点(1,f(x))处的切线垂直于直线x+2y+1=0.(1)求a,b的值.(2)若函数g(x)=ex/f(x),讨论g(x)的单调性.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 20:31:41
![设函数f(x)=ax2+bx+k(k>0)在x=0处取得极值,且曲线y=f(x)在点(1,f(x))处的切线垂直于直线x+2y+1=0.(1)求a,b的值.(2)若函数g(x)=ex/f(x),讨论g(x)的单调性.](/uploads/image/z/1969455-39-5.jpg?t=%26%23160%3B%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dax2%2Bbx%2Bk%28k%3E0%29%E5%9C%A8x%3D0%E5%A4%84%E5%8F%96%E5%BE%97%E6%9E%81%E5%80%BC%2C%E4%B8%94%E6%9B%B2%E7%BA%BFy%3Df%28x%29%E5%9C%A8%E7%82%B9%EF%BC%881%2Cf%28x%29%EF%BC%89%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E5%9E%82%E7%9B%B4%E4%BA%8E%E7%9B%B4%E7%BA%BFx%2B2y%2B1%3D0.%281%29%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC.%EF%BC%882%EF%BC%89%E8%8B%A5%E5%87%BD%E6%95%B0g%28x%29%3Dex%2Ff%28x%29%2C%E8%AE%A8%E8%AE%BAg%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E6%80%A7.)
xRN@v,mH/m",H+RB<%h^3-+~;meF}[چui>䈑if]ձs9'tUƐ_w IbqDޠ
,b۪7zkP%r4J}vڐZ9
ΡɆ0ZНoC^ mՆ -@~5b"YgKK-,nL[>$[FW=Ii>v- Oi5O$.$oEYe܄
设函数f(x)=ax2+bx+k(k>0)在x=0处取得极值,且曲线y=f(x)在点(1,f(x))处的切线垂直于直线x+2y+1=0.(1)求a,b的值.(2)若函数g(x)=ex/f(x),讨论g(x)的单调性.
设函数f(x)=ax2+bx+k(k>0)在x=0处取得极值,且曲线y=f(x)在点(1,f(x))处的切线垂直于直线x+2y+1=0.
(1)求a,b的值.
(2)若函数g(x)=ex/f(x),讨论g(x)的单调性.
设函数f(x)=ax2+bx+k(k>0)在x=0处取得极值,且曲线y=f(x)在点(1,f(x))处的切线垂直于直线x+2y+1=0.(1)求a,b的值.(2)若函数g(x)=ex/f(x),讨论g(x)的单调性.
(1)f(x)的导数为2ax+b,因为函数f(x)在x=0时取得极值,所以当x=0时,2ax+b=0.解得b=0.
因为曲线y=f(x)在点(1,f(x))处的切线垂直于直线x+2y+1=0,而直线x+2y+1=0的斜率为-0.5,则曲线y=f(x)在点(1,f(x))处的切线的斜率为2.所以当x=1时,2ax+b=2,解得a=1.
(2)由(1)得f(x)=x*x+k(k为正数)
将g(x)求导可知g(x)的导数值恒等于或者大于零.所以g(x)递增.