Question

在线急求，急急急急急，高中数学导数题，望高手解答，需过程，速给分

已知函数f(x)=(1/3)x的3次方-x的2次方+ax+b的图像在点p(0,f(0))处的切线是3x-y-2=0.求a,b的值.2设t属于(-2,-1),函数g(x)=f(x)+(m-3)x在(t,正无穷)上为增函数,求m的取值范围

Answer 1

f(0)=b,P（0,f(0))在3x-y-2=0上，f(0)=2,b=2.
f'(x)=x^2-2x+a,f'(0)=a,它等于3x-y-2=0的斜率，故a=3
a=3,b=2.
g(x)=(1/3)x^3-x^2+mx+2,
g'(x)=x^2-2x+m在(t,正无穷)上大于0，y=x^2-2x在x=1达最小值-1
故m>=1

Answer 2

1.f'(x)=x^2-2x+a为各点切线的斜率
则f'(0)=a=3
切线方程x=0时，y=-2，即(0,-2)在f(x)上，f(0)=-2，得b=-2

2.x>t时,g'(x)=f'(x)+(m-3)=x^2-2x+m=(x-1)^2+m-1>=0恒成立
而-2<t<-1<1,则g'(x)>=m-1,则m-1>=0,m>=1

Answer 3

1.f(0)=b,切点(0,b)在3x-y-2=0上，代入得b=-2,切线斜率3，f'(x)=x^2-2x+a,f'(1)=a-1=3,a=4.
2.当x>-2时g'(x)=f'(x)+m-3=x^2-2x+m+a-3=x^2-2x+m+1>=0，g'(x)=x^2-2x+m+1对称轴为x=1,
必有△=4-4(m+1)<=0,m>=0

Answer 4

x=0,f(0)=3x-2=-2=b,f(x)求导=x^2-2x+a,又令x=0则a=3 .................g(x)求导=x^2-2x+m=(x-1)^2+m-1 在(t,正无穷)上恒为正 则m-1恒大于等于0