Practice Problems
Section 3.3
1. (a) Cont. at 0, 1, 2, 3, 5 (b) Diff. at 0, 3, 5
3.
5.
(b) , the same result as in (a)
13. so Then implies that so and implies that
Finally, so implies that and . has , and
15. Their graphs are vertical shifts of each other, and their derivatives are equal.
21. so never equals 0 since never equals .
25. so for no values of (the discriminant .
27. so The graph of crosses the is infinitely often. The root of at is easy to see (and verify). Other roots of ', such as near and and , can be found numerically using the Bisection algorithm or graphically using the "zoom" or "trace" features on some calculators.
29. The graph of has two distinct "vertices" if for two distinct values of This occurs if the discriminant of is greater than
33. Everywhere except at and 3.
37. Everywhere. The only possible difficulty is at , and the definition of the derivative gives . The derivatives of the "two pieces" of match at to give a differentiable function there.
39. Continuity of at requires The "left derivative" at is and the "right derivative" of at is if then so to achieve differentiability and
(e) about seconds: up and down
(b) Max height when max height feet.
(b) -intercept at -intercept at
49. Since and are on the parabola, we need and . Then, subtracting the first equation from the second, .
so , the slope of Now solve the system and to get and Then use to get
(c) If for any constant , then .
53. (a) For so . Since , we know and .
(b) This graph is a vertical shift, up 1 unit, of the graph in part (a).