Answer
$\exists$ a real number x such that x(x + 1) > 0 and both x $\leq$ 0
and x $\geq$ −1.
Work Step by Step
Recall the form of the negation of a universal conditional statement:
$~(\forall x, P(x) \rightarrow Q(x)) \equiv \exists x$ such that $(P(x) \land $ ~$Q(x))$
Also recall De Morgan's Laws for the negation of OR:
~$(p \lor q) \equiv$ ~p $\land$ ~q
