Use the following pseudocode for the fixed point method to write MATLAB code to approximate a solution to x⁴ 3x² 3 = 0 on the interval [1, 2] with accuracy roughly within 10^-8 using x₀ = 1. Use maximum number of iteration of 100. Explain steps by commenting on them.

Algorithm : Bisection Method

Input: f(x)=e^x x2, interval [0,2], tolerance 10^-3, maximum number of iterations 50

Output: an approximate root of f on [0, 2] within 10^-3 or a message of failure

set a = 0 and b = 2;

xold = a; for i = 1 to 50 do set x = (a + b)/2; if |x xold| < 10^-3 then % checking required accuracy

FoundSolution = true; % done

Break; % leave for environment end if if f(a)f(x) < 0 then a = a and b = x

else a = x and b = b

end if xold = x; % update xold for the next iteration

end for if FoundSolution then

return x else print the required accuracy is not reached in 50 iterations end if

Algorithm : Fixed-point Iteration Input: g(x) = ln(x + 2), interval [0, 2], an initial approximation x₀, tolerance 10^-3, maximum number of iterations 50 Output: an approximate fixed point of g on [0, 2] within 10^-3 or a message of failure set x=x₀ and xold=x0;

for i = 1 to 50 do x = g(x); if |x xold| < 10^-3 then % checking required accuracy FoundSolution = true; % done break; end if xold = x; % update xold for the next iteration

end for if FoundSolution then

return x else print the required accuracy is not reached in 50 iterations

end if