*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **x − 3)2 (x + 2) x2 + 1 = 0. We get x = 0,
x = 3, and x = −2. (The equation x2 + 1 = 0 produces no real solutions.) These three points
divide the real number line into four intervals: (−∞, −2), (−2, 0), (0, 3) and (3, ∞). We select the
test values x = −3, x = −1, x = 1, and x = 4. We ﬁnd f (−3) is (+), f (−1) is (−) and f (1) is (+)
as is f (4). Wherever f is (+), its graph is above the x-axis; wherever f is (−), its graph is below
the x-axis. The x-intercepts of the graph of f are (−2, 0), (0, 0) and (3, 0). Knowing f is smooth
and continuous allows us to sketch its graph.
y (+) 0 (−) 0 (+) 0 (+)
−2
0
3
−3
−1
1 x 4
A sketch of y = f (x) A couple of notes about the Example 3.1.5 are in order. First, note that we purposefully did not
label the y -axis in the sketch of the graph of y = f (x). This is because the sign diagram gives us the
zeros and the relative position of the graph - it doesn’t give us any information as to how high or low
the graph strays from the...

View
Full
Document