**Unformatted text preview: **he category x < −1, we keep them all. For the second
case, we assume x ≥ −1. Our inequality becomes x + 1 ≥ x+4 , which gives 2x + 2 ≥ x + 4 or
2
x ≥ 2. Since all of these values of x are greater than or equal to −1, we accept all of these
solutions as well. Our ﬁnal answer is (−∞, −2] ∪ [2, ∞).
y
4
3
2 −4 −3 −2 −1 1 2 3 4 x We now turn our attention to quadratic inequalities. In the last example of Section 2.3, we needed
to determine the solution to x2 − x − 6 < 0. We will now re-visit this problem using some of the
techniques developed in this section not only to reinforce our solution in Section 2.3, but to also
help formulate a general analytic procedure for solving all quadratic inequalities. If we consider
f (x) = x2 − x − 6 and g (x) = 0, then solving x2 − x − 6 < 0 corresponds graphically to ﬁnding
the values of x for which the graph of y = f (x) = x2 − x − 6 (the parabola) is below the graph of
y = g (x) = 0 (the x-axis.) We’ve provided the graph again for referenc...

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