# Question #b2764

##### 1 Answer

Mar 4, 2017

#### Answer:

(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]}