Can you please solve this?
Answers
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:
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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