INSTRUCTOR 
Dr. Andrew Knyazev
Office: CU (Dravo) 620G. Phone: 556-8102.
Office hours: Tue 3pm - 6pm

NA I       EXAM #2     (Chapter 3)         October 16, 1995
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Part I. Bisection. Use initial interval [0,16] and equation

(x-1)(x-3)(x-5)(x-10)(x-12)=0.
                      


1. The method will find the root

[A]   1   [B]    3    [C]   5   [D]   10   [E]   12

2. The second approximation to the root is x =

[A]   4    [B]    6    [C]   8   [D]   10    [E]  None

3. To get estimate of the absolute error 1
one has to perform n iterations. Then n=

[A]   0    [B]    1   [C]   2   [D]    3   [E]   None
       
4. After 10 iterations of bisection the estimate
of the absolute error is

[A]   2   [B]    2   [C]   2   [D]  2     [E]   None




Part II. Newton method. 
 
5. Use initial guess  x = 0  and equation  (x-1)(x-3)=0.
Then the next approximation x  =

[A]  -.75  [B]    0   [C]   .75   [D]    1     [E]   None


6. Use initial guess  x = 0  and equation  

(x-1)(x-3)(x-5)(x-10)(x-12)=0. The method will find the root

[A]   1   [B]    3    [C]   5   [D]   10   [E]   None



7. Use initial guess  x = 1  and equation  

x(x+2)(x-2)=0. The method will find the root

[A]   -2   [B]    0   [C]   2   [D]     None
 

8. Use initial guess  x = 0  and equation 

(x-1)(x-3)(x+1)(x+3)=0. The method will find the root

[A]   -1   [B]    -3    [C]   1   [D]   3   [E]   None
 


Part III. Secant method. 

9. Use initial guesses  x = 0 and x = 1, and equation

(x-1)(x-3)=0. 

Then the next approximation x  =

[A]  -.75  [B]    0   [C]   .75   [D]    1     [E]   None


10. Use initial guesses  x = 0 and x = 1, and equation

(x+1)(x+3)=0. 

Then the next approximation x  =

[A]  -.75  [B]    0   [C]   .75   [D]    1     [E]   None