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Question 1
Let f(x) = (x-c) e-x+c – 0.3 where c = 3.558.
Using the simple iteration method, find the root of f(x) correct to TWO decimal places using x0= 2.0 + c. Hint: Check that the root satisfies x = ln(x-c) – ln 0.3  AND x = 0.3ex-c + c. You may need to iterate more than 5 times.
Question 2
Let f(x) = (x-c)2 e-x+c – 0.3 where c = 3.058.
Using the Newton-Raphson iteration scheme, find the root (correct to 3 decimal places) of f(x) with x0 = 1.0 + c.
Question 3
Let f(x) = (x-c) sin(x-c) + (x-c) + 5 where c = -0.022. If we set a0 = -6.7 + c, and b0 = -6.3+c and set (a0,b0) as the starting interval for the r=0 iteration of the bisection method. Find a4.
Question 4
Let f(x) = (x-c) e-x+c – 0.3 where c = 0.769. Using the simple iteration method, find the root of f(x) correct to TWO decimal places using x0= 0.4 + c.
Hint: Check that the root satisfies x = ln(x-c) – ln 0.3  AND x = 0.3ex-c + c. You may need to iterate more than 5 times.
Question 5
Letf(x) = a/x where a = 6.207. If R={x: 0.5 <= x  <= 1.0 }
Find the maximum value (correct to 3 decimal places) of| df/dx | in the interval of R.
Question 6
Let f(x) = eax + bx where a = 4.463 and b = 1.746.The Newton-Raphson method is used to solve f(x) = 0 with x0 = 0.67. Find x2 correct to 3 decimal places.
Question 7
We shall  let f(x) = ax2 where a = 0.98. If f(x) fulfills the condition 1 (Refer to SU1-8)of the contraction mapping in the interval
R = {x : 0 <= x <= c}
Find the maximum value of c. Express your answer correct to 3 decimal places.
Question 8
If f(x) = ax2 where a = 1.477.If | df/dx |  <=  1 in the interval R = { x:  0 <=  x <=  c }
find maximum c (correct to 3 decimal places).