MATH 180 FALL 1995


ASSIGNMENT #36


5.7, #1. f(x)=x3+3x2+3x-6, f¢(x)=3x2+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 xn+1=xn-(f(xn)/f¢(xn)) with x0=1. The truncated values for the next three xi 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. 3Ö{50} is a root of f(x)=x3-50. Since f¢(x)=3x2, 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 x0=3.5. The truncated values for the next three xi 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)=cosx-x, f¢(x)=-sinx-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)=-0.45969¼ < 0, so r is in (0,1). We start the iterative process with x0=1. The truncated values for the next three xi 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.


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