Question

The values of x which satisfy the inequality 1 / (x^3 + 3x^2) >= 0 are in the
interval(s)

(a) ] − 3,+1[
(b) [−3,+1[
(c) ] − 3, 0[ union [ ]0,+1[
(d) ]0, 3[ union ]3,+1[
(e) [+3,+1[

Answer 1

Testing "A" case:
]-3, +∞[ = {-2, -1, 0, 1, 2, 3, ...}

When x is equals 0, the equation has an indetermination.
$\frac{1}{x^3 + 3x^2} = \frac{1}{0^3 + 3 \cdot 0^2} = \frac{1}{0}$

Testing "B" case:
[-3, +∞[ = {-3, -2, -1, 0, 1, 2, 3, ...}

The same. When x is equals 0 or -3, the equation has an indetermination.
$\frac{1}{x^3 + 3x^2} = \frac{1}{(-3)^3 + 3 \cdot (-3)^2} = \frac{1}{0}$

Testing "C" case:
]-3, 0[ ∪ ]0, +∞[ = {-2, -1, 1, 2, 3, ...}
It's true.

For "D" and and "E", the range is incomplete.

Answer: Letter C