Consider the function fx x3 5x2 7x 3 Can you use the bis

Consider the function f(x) = x^3 - 5x^2 + 7x - 3 Can you use the bisection method to find the squareroot at x=1? Explain! If you use the bisection method, how many steps would you need at most to find the squareroot in the interval [2.3,4.4] with tolerance 10^-8? Demonstrate this in MATLABI Use the MATLAB fzero function to find the squareroots with starting intervals [0, 5], [2, 4] and [0, 2]. Plot the function, the line y=0 and the found squareroots in the interval [-1, 5] using the command scatter. Comment the results. In the textbook, page 161, there is an implementation of Newton-Raphsons method in MATLAB. Use this function to solve f(x) = 0 starting in x_0 = 0.5 and set the tolerance to 0, while enforcing maximum number of iterations to 6. Make the method tell the value of the guess in each iteration with 10 digits precision. Compare f zero and newtraph. Make a table of iterations. Does f zero give results that converge quadratically?

Solution

f(0)=-3,f(1)=0;\\\\

(a).x=1 is one of the root of the given equation.

(b )a=2.3;f(a)=14.6341
b=4.4;f(b)=305.8096
x1=(a+b)/2;f(x1)=90.2820