MATH 180 FALL 1995

ASSIGNMENT #36

5.7, #1. f(x) = x^{3} + 3x^{2} + 3x − 6, f′(x) = 3x^{2} + 6x + 3 = 3(x + 1)^{2}. In particular, f′(x) is never negative so that f(x) is increasing. Thus f has only one root r. Since f(0) = −6, f(1) = 1, it follows that r is in the interval (0, 1). We start the iterative process x_{n+1} = x_{n} − (f(x_{n})/f′(x_{n})) with x_{0} = 1. The truncated values for the next three x_{i} are

0.9166667 0.9129384 0.9129311

A calculation shows that f(0.9129311) > 0 and f(0.9129300) < 0. The sign change shows that 0.9129300 < r < 0.9129311. Thus 0.913 approximates r to 2 decimal places since |0.913 − r| < 0.913 − 0.9129300 < 1/200.

5.6, #2. ^{3} \sqrt{50} is a root of f(x) = x^{3} − 50. Since f′(x) = 3x^{2}, f(x) is increasing and f has only one root r. Now f(4) = 14 > 0, f(3.5) = −7.125 < 0, so r is in (3.5, 4). We start the iterative process with x_{0} = 3.5. The truncated values for the next three x_{i} are

3.693877 3.684057 3.684031

A calculation shows that f(3.68406) > 0 and f(3.68403) < 0. The sign change shows that 3.68403 < r < 3.68406. Thus 3.684 approximates r to 2 decimal places since |3.684 − r| < 3.68406 − 3.684 < 1/200.

5.6, #6. f(x) = cos x − x, f′(x) = − sin x − 1. In particular, f′(x) is never positive. So f(x) is decreasing and f has only one root r. Now f(0) = 1 > 0, f(1) = −0.45969 ⋯ < 0, so r is in (0, 1). We start the iterative process with x_{0} = 1. The truncated values for the next three x_{i} are

0.750363 0.739112 0.739085

A calculation shows that f(0.73908) > 0 and f(0.73911) < 0. The sign change shows that 0.73908 < r < 0.73911. Thus 0.739 approximates r to 2 decimal places since |0.739 − r| < 0.73911 − 0.739 < 1/200.