已知曲线y=x^2+ax+b与2y=xy^3-1在点（1，-1）处有公共切线，求常数a,b
最新回答
一川绿风
2024-11-04 13:43:10
简单计算一下即可，答案如图所示
容嬷嬷的春天
2024-11-04 11:04:46
如图
风夏了白雪
2024-11-04 02:11:33
请
微浅笑°
2024-11-04 10:57:20
y=x^2+ax+b
-1=1+a+b
a+b=-2 (1)
y=x^2+ax+b
y' = 2x+a
y'|(1,-1) = 2+a
2y=xy^3-1
2y' = 3xy^2.y' + y^3
y' = y^3/(2-3xy^2)
y'|(1,-1) = -1/(2-3) = 1
2+a = 1
a=-1
from (1)
a+b=-2
-1+b=-2
b=-1
(a,b)=(-1,-1)
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