f(x) = (x-c) e

^{-x+c}– 0.3 where c = 3.558.

TWO decimal places using x

_{0}= 2.0 + c.

Hint: Check that the root satisfies x = ln(x-c) – ln 0.3 AND x = 0.3e^{x-c} + c. You may need to iterate more than 5 times.

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 x

_{0 }= 1.0 + c.

f(x) = (x-c) sin(x-c) + (x-c) + 5

where c = -0.022.If we set

a_{0} = -6.7 + c, and b_{0} = -6.3+c and set (a_{0},b_{0}) as the

starting interval for the r=0 iteration of the bisection method.

Find a_{4}.

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 x

_{0}= 0.4 + c.

Hint: Check that the root satisfies x = ln(x-c) – ln 0.3 AND x = 0.3e^{x-c} + c. You may need to iterate more than 5 times.

f(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.

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f(x) = e

^{ax}+ bx

where a = 4.463 and b = 1.746.

The Newton-Raphson method is used to solve f(x) = 0 with x_{0} = 0.67.

Find x_{2} correct to 3 decimal places.

f(x) = ax

^{2}where a = 0.98. If f(x) fulfills the condition 1 (Refer to SU1-8)

of the contraction mapping in the interval

your answer correct to 3 decimal places.

f(x) = ax

^{2}

where a = 1.477.

If

| df/dx | <= 1

in the interval

R = { x: 0 <= x <= c }

find maximum c (correct to 3 decimal places).

f(x) = 0 In the 0th iteration (the zeroth iteration, i.e., r = 0), an

interval of

a <= x <= b is used,

Find the length of the interval in the third iteration (i.e., r = 3).

Give the answer correct to 3 decimal places.

Question 10

f(x) = (x-c)

^{2}e

^{-x+c}– 0.3 where c = 3.549.

Using the Newton-Raphson iteration scheme, find the root (correct to 3 decimal places) of f(x) with x

_{0 }= 4.0 + c.