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n this module, I include a file called Bisection Technique. I use an example similar to the textbook described in section 2.1.  I start to solve the example by hand and complete it using MATLAB. After reading the Bisection technique lesson 4 and section 2.1 of the textbook, answer the following discussion question in your own words.When running the Bisection method in lesson 4 (program 1.1), with a tolerance of 0.001 the answer is 1.3652 which is equivalent to p9 according to the table 2.1 from the textbook.  When running p13 in lesson 4 (program 1.2), the answer is 1.3651 which is equivalent to p13.  Which one of the answers do you think is the most accurate answer closest to the solution and why? Which of the two calculation methods do you prefer and why?  Elaborate in your answers.file attachedLesson 4Bisection TechniqueTo find a solution or root of an equation f(x) we can apply the bisection method, which is anapproximation technique to get closer and closer to the value of the root by dividing theinterval in half after each iteration.  The bisection method or binary search technique is basedon the Intermediate Value Theorem.  Suppose f is a continuous function on the interval [a,b]with f(a) and f(b) of opposite sign, the Intermediate Value Theorem implies that a number pexists in [a,b] with f(p) = 0.  The method calls for a repeated halving or bisecting of subintervalsof [a,b] and at each step locating the half containing p.Example:Doesf(x) = x3+4×2-10has a root or solution in[a,b] = [1,2]?1)   Let’s check if f(1) and f(2) have opposite signs according to the Intermediate ValueTheorem.f(a) = f(1)= (1)3+ 4(1)2– 10 = -5(negative)f(b) = f(2)= (2)3+ 4(2)2– 10 = 8+16-10 = 14(positive)The answer is yes, which means since f is continuous there must be a solution in theinterval [1,2].2)   Let p be the solution we are searching for and let p1 be the first midpoint of [a,b] = [1,2].p1= (a+b)/2 = (1+2)/2 = 3/2 = 1.5f(p1)=f(1.5)= (1.5)3+4(1.5)2-10 = 2.375(positive)3)   Iff(p1)= 0then p1 is the root or the solution, so p= p1. Done.Otherwise, iff(p1)is not equal to zero, which is the case, we’re going to replace either a or b byp1 to obtain a new interval closer to the root or the solution.How do we know which one to replace?  Simple.  The goal is to keep two intervals of whichtheir functions have opposite signs when we calculate the functionfat these two intervals.  Thequestion to ask is rather which of the twof(a)andf(b)has opposite sign to the functionf(p1)?a)   Iff(p1)andf(a)have the same sign, then they don’t satisfy the Intermediate ValueTheorem.  We replaceabyp1, so the new interval that contains the solution p will be[p1, b].b)   Iff(p1)andf(a)have opposite signs, then they satisfy the Intermediate Value Theorem.We replacebbyp1so the new interval that contains the solution p will be [a, p1].In our case,f(a) which is f(1)has opposite sign tof(p1) which is f(1.5). We replace (bwhich is 2 byp1which is 1.5).  The new interval that contains a solution is [a, p1] = [1,1.5].Next we go back to step 2 and step 3 by repeating this process again to find the newmidpoint p2, check if p2 is the solution, if not obtain a new interval.Let’s continue with steps 2 and 3.Step 2:a.   let p2 be the second midpoint of [a, p1] = [1, 1.5].p2= (a+p1)/2 = (1+1.5)/2 = 2.5/2 = 1.25f(p2)= f(1.25) = (1.25)3+4(1.25)2-10 = -1.796875(negative)Step 3:Which of the twof(a)andf(p1)has opposite sign to the functionf(p2)ORWhich of the twof(1)andf(1.5)has opposite sign to the functionf(1.25)?f(a)=f(1)= -5;f(p1) = f(1.5) = 2.375;f(p2) = f(1.25) = -1.796875The answer isf(p1) = f(1.5)andf(p2) = f(1.25), so we replace a by p2.Therefore, the new interval that contains a solution is: [p2, p1] = [1.25, 1.5].We repeat steps 2 and 3 again and again untilf(midpoint) = 0or within a Tolerance, let’ssay 0.001 = 10-4.To do this we can use the MATLAB tool.  The following program 1.1 can be used to findthe root of the same function using bisection method with a tolerance of 0.001 = 10-4%Program 1.1 Bisection Method%Computes approximate solution of f(x)=0%Input: inline function f; a,b such that f(a)*f(b)<0,%     and tolerance tol%Output: Approximate solution xcfunction xc = bisect(f,a,b,tol)if sign(f(a))*sign(f(b)) >= 0,error(‘f(a)f(b)<0 not satisfied!')%ceases executionendfa=f(a);fb=f(b);k = 0;while (b-a)/2>tolc=(a+b)/2;fc=f(c);if fc == 0%c is a solution, donebreakendif sign(fc)*sign(fa)<0  %a and c make the new intervalb=c;fb=fc;else                    %c and b make the new intervala=c;fa=fc;endendxc=(a+b)/2;%new midpoint is best estimate1.  Copy and paste this program in MATLAB in the editor window.2.  Save the program as bisect.3.  To run the program, type the following in the command window:>> bisect(@(x) x^3+4*x^2-10,1,2,0.0001)@(x) x^3+4*x^2-10represents the function1,2represents the interval0.01represents the Tolerance.The output is:ans =1.3652The following MATLAB program 1.2 can be used to find the root of the same functionusing bisection method with 13 iterations meaning that we will find the midpoint 13 times pn =p13.%Program 1.2 Bisection Methodfunction [x e] = mybisect(f,a,b,n)% function [x e] = mybisect(f,a,b,n)% Does n iterations of the bisection method for a function f% Inputs: f — an inline function%         a,b — left and right edges of the interval%         n — the number of bisections to do.% Outputs: x — the estimated solution of f(x) = 0%          e — an upper bound on the errorformat longc = f(a); d = f(b);if c*d > 0.0error(‘Function has same sign at both endpoints.’)enddisp(‘           x                  y’)for i = 1:nx = (a + b)/2;y = f(x);disp([    x     y])if y == 0.0% solved the equation exactlye = 0;break% jumps out of the for loopendif c*y < 0b=x;elsea=x;endende = (b-a)/2;1.  Copy and paste this program in MATLAB in the editor window.2.  Save the program as bisection.3.  To run the program, type the following in the command window:>> bisection(@(x) x^3+4*x^2-10,1,2,13)@(x) x^3+4*x^2-10represents the function1,2represents the interval13represents the number of iterations.The output is:xy1.500000000000000   2.3750000000000001.250000000000000  -1.7968750000000001.375000000000000   0.1621093750000001.312500000000000  -0.8483886718750001.343750000000000  -0.3509826660156251.359375000000000  -0.0964088439941411.367187500000000   0.0323557853698731.363281250000000  -0.0321499705314641.365234375000000   0.0000720247626301.364257812500000  -0.0160466907545921.364746093750000  -0.0079892628127711.364990234375000  -0.0039591015229231.365112304687500  -0.001943659010067ans =1.365112304687500

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