Question #b2764
1 Answer
Mar 4, 2017
(c) 2 and 3 only
Explanation:
Given:
#f(x) = x^(-1/3)#
Note that:
#f(0) = 1/0" "# is undefined
#lim_(x->0+) x^(-1/3) = lim_(t->0+) 1/t = oo#
#lim_(x->0-) x^(-1/3) = lim_(t->0-) 1/t = -oo#
So 1. is false and 2. is true.
In order to check 3., we need to split the integral into two parts, noting that:
#f(x) < 0" "# when#x in [-1, 0)#
#f(x) > 0" "# when#x in [1, 0)#
So:
#A = abs(int_(-1)^0 x^(-1/3) dx) + abs(int_0^1 x^(-1/3) dx)#
#color(white)(A) = abs([3/2 x^(2/3)]_(-1)^0) + abs([3/2 x^(2/3)]_0^1])#
#color(white)(A) = abs(0-3/2) + abs(3/2 - 0)#
#color(white)(A) = 3/2+3/2#
#color(white)(A) = 3#
So the area is finite and 3. is true.
graph{(y-x^(-1/3))(x+1+0.00001y)(x-1+0.00001y) = 0 [-5.086, 4.92, -2.55, 2.45]}