Stitz-Zeager_College_Algebra_e-book

This sort of behavior reminds us of parabolas and

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x2 ≥ |x − 1| − 1, we re-write the absolute value using cases. For x < 1, |x − 1| = −(x − 1) = 1 − x, so we get 2x − x2 ≥ 1 − x − 1, or x2 − 3x ≤ 0. Finding the zeros of f (x) = x2 − 3x, we get x = 0 and x = 3. However, we are only concerned with the portion of the number line where x < 1, so the only zero that we concern ourselves with is x = 0. This divides the interval x < 1 into two intervals: (−∞, 0) and (0, 1). We 1 1 choose x = −1 and x = 2 as our test values. We find f (−1) = 4 and f 2 = − 5 . Solving 4 2 − 3x ≤ 0 for x < 1 gives us [0, 1). Next, we turn our attention to the case x ≥ 1. Here, x |x − 1| = x − 1, so our original inequality becomes 2x − x2 ≥ x − 1 − 1, or x2 − x − 2 ≤ 0. Setting g (x) = x2 − x − 2, we find the zeros of g to be x = −1 and x = 2. Of these, only x = 2 lies in the region x ≥ 1, so we ignore x = −1. Our test intervals are now [1, 2) and (2, ∞). We choose x = 1 and x = 3 as our test values and find g (1) = −2 and g (3) = 4. To solve g (x) ≤ 0, we have [...
View Full Document

Ask a homework question - tutors are online