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1. To solve f (x) = g (x), we look for where the graphs of f and g intersect. These appear to be
at the points (−1, 2) and (1, 2), so our solutions to f (x) = g (x) are x = −1 and x = 1.
2. To solve f (x) < g (x), we look for where the graph of f is below the graph of g . This appears
to happen for the x values less than −1 and greater than 1. Our solution is (−∞, −1) ∪ (1, ∞).
3. To solve f (x) ≥ g (x), we look for solutions to f (x) = g (x) as well as f (x) > g (x). We solved
the former equation and found x = ±1. To solve f (x) > g (x), we look for where the graph
of f is above the graph of g . This appears to happen between x = −1 and x = 1, on the
interval (−1, 1). Hence, our solution to f (x) ≥ g (x) is [−1, 1].
y y 4 4
y = g (x) 3
(−1, 2) (−1, 2) (1, 2) 2 −1 (1, 2) 2 1
−2 y = g (x) 3 1
1 2 x −1 −2 −1 1 y = f (x)
f (x) < g (x) 2 x −1
y = f (x)
f (x) ≥ g (x) 2.4 Inequalities 157 We now turn our attention to solving inequalities involving the absolute value. We have the
following theorem from Intermediate Algebra to help us.
Theorem 2.3. Inequalities Involving the Absolute Value: Let c be a real number.
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