Approximate a root of function surfaceAreaPrime given its first derivative SurfaceAreaDoublePrime, an initial guess (xGuess) and an error limit.

Question

What will be an ideal response?

Answer 1

```
double newton (double xGuess, double okError, int& converges)
{
double xN, xNplus1, absDiff;
int iter = 0;

xN = xGuess;
do {
++iter;
xNplus1 = xN -
surfaceAreaPrime(xN) / surfaceAreaDoublePrime(xN);
absDiff = fabs(xN-xNplus1);
xN = xNplus1;
} while (absDiff >= okError && iter < MAX_ITER);

if (absDiff < okError) {
converges = 1;
return xN;
} else {
cout << endl << "Newton's method did not converge to a "
<< "minimum surface area in " << MAX_ITER << endl
<< "iterations using an initial guess of " << xGuess
<< " centimeters." << endl;
converges = 0;
return xGuess;
}
}

double surfaceArea (double r)
{
return (2 * M_PI * r * r + 4000 / r);
}

double surfaceAreaPrime (double r)
{
return (4 * M_PI * r - 4000 / (r * r));
}

double surfaceAreaDoublePrime (double r)
{
return (4 * M_PI + 8000 / (r * r * r));
}

```