Question

求连续区间f(x)=(x^3+3x^2-x-3)/(x^2+x-6)

Answer 1

f(x)=(x^3+3x^2-x-3)/(x^2+x-6)
=(x+1)(x-1)(x+3)/[(x-2)(x+3)]
=(x+1)(x-1)/(x-2)
求一阶导数 y'=(x^2-4x)/(x-2)=0
x=0或x=4
x<0，且x不=-3时 f(x)递增
0<=x<4，且x不=2时 f(x)递减
x>=4时 f(x)递增

Answer 2

f(x)=x^2(x+3)-(x+3)/(x+3)(x-2)=(x+3)(x^2-1)/(x+3)(x-2)=(x+1)(x-1)/(x-2)

初等函数连续,x≠2

连续区间(-∞,2)∪(2,+∞)

Answer 3

将分子分母分解因式，其中分母为(x+2)(x-3)，所以3和-2点为函数间断点，再令(x+1)(x-1)(x+3)=k(x+2)(x-3)验证，最后确定区间

Answer 4

(-∞,2)∪(2,+∞)