Question

曲线x3-3x2+1在点（1,-1)处的切线方程为 A.y=-3x+2 B.y=3x-4 C.y=-4x+3 D.y=4x-5

Answer 1

这个就是切点
y'=3x2-6x
x=1,斜率k=y'=-3
所以是3x+y-2=0

Answer 2

y=-(x-1)/(2x+5)
=-(x+2.5-3.5)/[2(x+2.5)]
=-(x+2.5)/[2(x+2.5)]+3.5/[2(x+2.5)]
=-1/2+7/(4x+10)
即求7/(4x+10)减区间
定义域4x+10≠0
x≠-5/2
分子7>0,所以x<-5/2和x>-5/2都递减
所以减区间是(-∞,-5/2)和(-5/2,+∞)