Question

设函数f(x)=x^3+ax^2-9x-1(a<0)若曲线y=f(x)的斜率最小的切线与直线y=-1

设函数f(x)=x^3+ax^2-9x-1(a<0)若曲线y=f(x)的斜率最小的切线与直线y=-12x+6平行，(1)球a值；(2)函数f(x)的单调曲间

Answer 1

f'(x)=3x^2+2ax-9=3(x+a/3)^2-9-a^2/3
当x=-a/3时，f'(x)有最小值-9-a^2/3，切线有最小斜率-9-a^2/3。又此切线与已知直线平行，斜率相等
-9-a^2/3=-12
a^2=9
a=3(a<0，舍去)或a=-3
a=-3
f'(x)=3x^2-6x-9=3(x-1)^2-12
令f'(x)≥0 3(x-1)^2-12≥0 (x-1)^2≥4 x≥3或x≤-1
令f'(x)≤0 3(x-1)^2-12≤0 (x-1)^2≤4 -1≤x≤3
函数的单调递增区间为(-无穷,-1]U[3，+无穷)，单调递减区间为[-1，3]

Answer 2

f'(x)=3x^2+2ax-9=3(x+a/3)^2-9-a^2/3
当x=-a/3时，f'(x)有最小值-9-a^2/3，切线有最小斜率-9-a^2/3。又此切线与已知直线平行，斜率相等
-9-a^2/3=-12
a^2=9
a=3(a<0，舍去)或a=-3
a=-3
f'(x)=3x^2-6x-9=3(x-1)^2-12
令f'(x)≥0 3(x-1)^2-12≥0 (x-1)^2≥4 x≥3或x≤-1
令f'(x)≤0 3(x-1)^2-12≤0 (x-1)^2≤4 -1≤x≤3
函数的单调递增区间为(-无穷,-1]U[3，+无穷)，单调递减区间为[-1，3]
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