Question

Can you please solve this?

Answer 1

Java program:

// Bisection Method Solves: x^2 - 3 = 0

public class Bisection01

{

public static void main(String[ ] args)

{

final double epsilon = 0.00001;

double a, b, a, y_m, y_a:

a = 0; b = 4:

while ( (b-a) epsilon )

{

m= (a+b)/2: // Mid point

y_m = m*m - 3.0; 11 y_m = f(n)

y_a = a*a - 3.0; // ya f(a)

if ( (y_m >0 && ya < 0) || (y_m <0 && y# > 0) )

{ // f(a) and f(m) have different signs: move b b = m;

}

else

{ // f(a) and f(n) have same signs: move a

a = m;

}

System.out.println("New Interval: [ " + a + " . . " + b + "] ");

// Print progress

System.out.println("Approximate solution = " + (a+b)/2 );

}

}

Output: see the attachment

Answer 2

Answer:

0.02

Step-by-step explanation:

For solution we always take the lower interval number that is 1

and dy = f'(x) . δx

=> dy = (2x) . (0.01)

=> dy = 2 (1) (0.01)

=> dy = 0.02

Hope it helps

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