Question

Question #b2764

Answer 1

(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]}