Question

函数连续和间断点咋求？麻烦写下步骤

Answer 1

(1)

f(x)
=1 ; x<-1
=x ;-1≤x≤1
=1 ; x>1
f(-1-)=lim(x->-1-) 1 = 1
f(-1)=f(-1+)lim(x->-1+) x = -1≠f(-1-)
x=-1, f(x) 不连续
f(1)=f1-)=lim(x->1)x = 1
f(1+)=lim(x->1+)1 =1
x=1, f(x) 连续
(2)
(1)
g(x)=x^3+3x^2-x-3
g(1) =0
x^3+3x^2-x-3 =(x-1)(x^2+ax+3)
coef. of x
3-a=-1
a=4
x^3+3x^2-x-3
=(x-1)(x^2+4x+3)
=(x-1)(x+3)(x+1)
f(x)

=(x^3+3x^2-x-3)/(x^2+x-6)
=(x-1)(x+3)(x+1)/[(x+3)(x-2)]
x=1 , x=-1 ；无穷间断点
x=-3 ； 可去间断点
连续区域 =(-∞, -3) U (-3, -1) U (-1, 1) U (1,+∞)
(2)

f(x)
=1/x ; x<0
=x^2 ; 0≤x≤1
=2x-1 ; x>1
f(0-)= lim(x->0-)1/x ->-∞
f(0) =f(0+)=lim(x->0+) x^2 =0
x=0 ; 无穷间断点
f(1)=f(1-)=lim(x->1-) x^2 =1
f(1+)=lim(x->1+) (2x-1) = 2-1 =1
x=1, f(x) 连续
连续区域 = (-∞, 0) U (0,+∞)

Answer 2