Question

1.设y=2x^3-12x^2+a在[-1,2]上最大值为2,求a 2.求由方程ycosx+sin（x-y）=0确定的隐

Answer 1

1、y=2x³-12x²+a

求导 y'=6x²-24x

令y'=0 解得 x=0 或 x=4

∴y在x∈[-1,2]上先增后减,在x=0处取得最大值

∴y(x=0)=a=2

∴a=2

2、ycosx+sin（x-y）=0

求导得：y'cosx-ysinx+cos(x-y)*(1-y')=0

y'[cosx-cos(x-y)]=ysinx-cos(x-y)

y'=[ysinx-cos(x-y)]/[cosx-cos(x-y)]