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(Numerical analysis) Heres a challenging problem for those who know a little calculus. The Newton-Raphson method can be used to find the roots of any equation0.n this method the (i + Dst approximation, xi-1, to a root ofy(x) = 0 İs given in terms of the ith approximation, xi by the following formula, where y denotes the derivative of y(x) with respect tox: For example, if yx) 32+2x - 2, then y()-2, and the roots are found by making a reasonable guess for a first approximation x and iterating by using this equation: a. Using the Newton-Raphson method, find the two roots of the equation 32+2-2 0. Hint: Theres one positive root and one negative root.) b. Extend the program written for Exercise 6a so that it finds the roots of any function (x)-0, when the function for y(x) and the derivative of y(x) are placed in the code.

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is program evaluates the value of roots of the equation given below: 3x2 +2x-2-0 The roots of the equation are evaluated as the expression given below x-x,-(3x, +2x,-2)/(6x +2) where, x41 is the new approximate value of x and x, is the previous approximate value of x . The value of root is evaluated until the difference between two consecutive approximate values becomes lesser than a very small difference such as 0-000001Program Plan: Include header files and declare required namespaces. Iniialize the main () function. . Declare variables for previous, current, and temporary values of root and the minimum differences between consecutive values of root. Implement the first do-while loop to execute until difference between positive consecutive root approximations reaches minDiff. Evaluate next approximation for each execution of loop. Implement the second do-while loop to execute until difference between negative consecutive root approximations reaches minDiff. Evaluate next r each execution of 1 .Check exit conditions to either exit both loops or update root values. .Display formatted output and end program after loop exits.k*kk* k*k*k k* // Program to compute approximate value of root of number // until difference between approximations is 0.000001 // Header files #include <iostream> #include ciomanip> #include <cmath> /Namespace use using namespace std: /Initialize main) function int main() We declare double variables xPositivePrevious (previous positive root value; initialized to 1.0 as guess), xPositiveNext (current positive root value), xNegativePrevious (previous positive root value; initialized to -1.0 as guess) xNegativeNext (current positive root value), minDiff (minimum difference value specified), tempPositive (temporary previous positive value), and tempNegative (temporary previous negative value). // declare variables double xPositivePrevious1.0, xNegativePrevious --1.0, XNegativeNext, mìnDiff 0.000001, tempPositive, tempNegative: xPositiveNext,We loop the program continuously to evaluate the positive root of the equation. Initially, we set the root to a guess of 1.0, and then evaluate subsequent root values. If the difference between two consecutive root approximations is less than 0.000001 ( very small value), we break loop and exit. // loop until difference between two values is less than minDiff do // evaluate and update to new value xpositiveNext xpositivePrevious- ( ( 3 * pow ( xPositivePrevious, 2 2 * xPositivePrevious )-2 6xPositivePrevious 2 /7 check for difference in values to be lesser than minDiff if abs(xPositivePrevious - xPositiveNext minDiff // if true, exit loop break; // store temporary previous positive value temppositive xpositivePrevious ; // update previous positive value to current value XPositivePrevious = XPositiveNext ;/I loop continuously while true We loop the program continuously to evaluate the positive root of the equation Initially, we set the root to a guess of -1.0 and then evaluate subsequent root values. If the difference between two consecutive root approximations is less than 0.000001 (very small value), we break loop and exit. // loop until difference between two values is less than minDiff // evaluate and update to new value xNegativeNextxNegativePrevious-3 pow xNegativePrevious, 2 2 * xNegativePrevious ) - 2 /6xNegativePrevious 2 // check for difference in values to be lesser than minDiff if abs xNegativePreviousXNegativeNext < minDiff) // if true, exit loop break; We set the temporary negative root to store the previous root value and update the previous root value to the current root for the next execution of the loop// store temporary previous negative value tempNegative = XNegative Previous ; // update previous negative value to current value xNegativePrevious = XNegativeNext ; / loop continuously while true We implement the setprecision) method of the iomanip header to set the precision of floating point values to 6. We also update the previous positive and negative values of x to the actual previous values and display output. /I update positive and negative values to status before the final execution again XPositivePrevious tempPositive ; xNegativePrevious - tempNegative; // Display positive root as output cout<< setprecision 6) << fixed<<For the minimum difference << minDiff << , the approximate positive root is<xPositiveNext <<fo previous approximation << xPositivePrevious << endl << endl; // Display negative root as output cout << setprecision 6 << fixed << For the minimum difference << minDiff << , the approximate negative root is <<xNegativeNext <<for previous approximation << xNegativePrevious;Sample Output: For the minimum difference 0.000001, the approximate positive root is 0.548584 for previous approximation 0.548589 For the minimum difference 0.000001, th approximate negative root is -1.215250 for previous approximation- 1.215909

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