*Answer Quick within 25mins thanks*

Use the code below to answer this question. % Newton Raphson iteration to find roots of cos(x)*cosh(x) = -1 clc:clear, X=...% Enter an initial guess here fx = cos(x)* cos(x) + 1: while abs(fx) > 0.001 fprintf(x =%6.3f f(x) =%8.41 ', x, fx) dfdx = -sin(x) cosh(x) + cos(x)" sinh(x); * = x - fx/dfdx: fx = cos(x) * cosh(x) + 1; end fprintf('=%6.3f f(x) = %8.4f ', X, fx): What would the output table look like? You do not need to calculate any values to answer this question. None of these answers Ox=4.500 f(x) = -8.488 X = 4.746 f(x) = 2.9307 * = 4.696 f(x) = 0.1375 Ox=4.500 f(x) = -8.4888 X = 4.746 f(x) = 2.9307 X = 4.696 f(x) = 1.1375 X = 4.694 f(x) - 1.1014 Ox=4.500 F(x) = -8.4888 x = 4.746 f(x) = 2.9307 x = 4.696 f(x) - 0.1375 X= 4.694 f(x) = 0.0004 Ox=4.500 f(x) = -8.4888 X = 4.746 f(x) = 2.9307 X= 4.696 f(x) = 0.1375