Question

write me a java progrm that does this:

y=x^3 +1

drivative of y=3X^2+0

y=mx+b

m is slope for x=2 put 2 back to equation y=3(2)^2=12

now we have 12 and y=9 so

9=2(12)+b and we find b=-15

now we write new equation that with these information

y=12x-15 we solve for 12x-15=0, we solve for x and x=15/12=1.25

now we do the same thing above we put at the drivative equation y=3(1.25)^2+0=4.6875

now we put this to 2.943125=(4.6875)*3+b

we solve for b=-2.90625

y=4.6875x-2.90625 and 4.6875x-2.90625=0 we solve for x and =0.62 now we use this as we did above to find another new equation and this go in a loop to do the same thing over and over to find new point and equation

Answer 1

/*----------------------------------------------------\
|save code 'equation.java' |
|compile using 'javac equation.java'       |
|run using 'java equation' |
\----------------------------------------------------*/

import java.lang.Math;

//your file name class name must be same here file name is 'equation.java'
class equation{
   public static void main(String args[]){

       float x = 2f;//initial value of x which is given
       int i = 6;//loop iterator
       for(i=0; i<6; i++){//calling in loop over and over again
           x = printEquation(x);
       }

   }
   // Function to find values and equation
   public static float printEquation(float x){
       float y1, m, b, output;
       y1 = (float) Math.pow(x, 3) + 1;//given function---(1)
       m = (float) Math.pow(x, 2) * 3;//darivative of function(1) slope
       b = (float) y1 - m*x;//intersection part of equation
       output = (float) (-1)*b / m;//next x value
       System.out.println("\nFor value x = "+x);
       System.out.println("Equation will be y = ("+m+")*x + ("+b+")\n");
       return output;
   }
}

//screen shots