0
$\begingroup$

Write $A \cap B$ as the union of intervals. Use standard interval notation, that may include round brackets and square brackets. None of your intervals should touch or intersect another.

$\begin{align*} A &= \Bigl\{ x\in\mathbb{R}\Bigm| x^2 + 3x - 4 \gt 0\Bigr\}\\ B &= \Bigl\{ x\in\mathbb{R}\Bigm| x^2 + 3x - 10\leq 0\Bigr\} \end{align*}$

  • 1
    Do you know how to figure out what *each* of $A$ and $B$ are as (unions of) intervals? HINT: Factor the quadratics.2012-03-07

1 Answers 1

1

In case this is a homework, here is a hint on a similar example.

Let $ A = \{ x \in \mathbb{R} | x^2-x-6 > 0 \}.$ First, factor $ x^2-x-6 = (x-3)(x+2) = 0$. Hence there are two roots at $x =3, x= -2$. This divides up the number line $\mathbb{R}$ into three intervals: $ (-\infty, -2), (-2, 3), (3, \infty). $ Now you need to check the sign of $(x-3)(x+2)$ in each of the three intervals. Only $x \in (3, \infty) \cup (-\infty, -2)$ makes $(x-3)(x+2)$ positive, i.e. $x^2-x-6 > 0.$ Hence $ A = (3, \infty) \cup (-\infty, -2). $ Do the same with $A,B$ in your question. Then $A \cup B$ should be apparent. Note: pay attention that $B$ has $\le$ in the definition, so some intervals will be closed (rather than open).

  • 0
    @ArturoMagidin Ops. I'm sorry. I did not mean to accuse the OP. I just want to help without crossing the line of `(homework)`. I'll edit my answer.2012-03-07