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1.2 Answers for Numerical Methods 633

ANSWERS FOR NUMERICAL METHODS
Exercise Set 1.2 (Page 000)
1. For each part, f ¡ô C[a, b] on the given interval. Since f(a) and f(b) are of opposite
sign, the Intermediate Value Theorem implies a number c exists with f(c) ¡ë 0.
3. For each part, f ¡ô C[a, b], f
exists on (a, b), and f(a) ¡ë f(b) ¡ë 0. Rolles Theorem
implies that a number c exists in (a, b) with f
(c) ¡ë 0. For part (d), we can use
[a, b] ¡ë [−1, 0] or [a, b] ¡ë [0, 2].
5. a. P2(x) ¡ë 0

b. R2(0.5) ¡ë 0.125; actual error ¡ë 0.125

c. P2(x) ¡ë 1+3(x − 1) + 3(x − 1)2

d. R2(0.5) ¡ë −0.125; actual error ¡ë −0.125

7. Since
P2(x) ¡ë 1+x and R2(x) ¡ë
−2e¥î(sin ¥î + cos ¥î)
6 x3

for some number ¥î between x and 0, we have the following:
a. P2(0.5) ¡ë 1.5 and f(0.5) ¡ë 1.446889. An error bound is 0.093222 and |f(0.5)−
P2(0.5)| ¡Â 0.0532
b. |f(x) − P2(x)| ¡Â 1.252
c.

1
0 f(x) dx ≈ 1.5
d. |

1
0 f(x) dx−

1
0 P2(x) dx| ¡Â

1
0
|R2(x)| dx ¡Â 0.313, and the actual error is 0.122.
9. The erro