clc; clear

$\%$ Root lies between in the interval $[$a$,$b$].$

f $=$ input$(\text{'}$Enter the function f$($x$)$: $\text{'})$;

$\%$ To input a function you can use either of these two methods:

$\%$ Method $1$: @$($x$)$ sqrt$(9\mathrm{.}81*$x$/0\mathrm{.}25)0\mathrm{.}25/$x$)*4)-36$

$\%$ Method $2$: inline$(\text{'}$sqrt$(9\mathrm{.}81*$x$/0\mathrm{.}25)0\mathrm{.}25/$x$)*4)-36\text{'})$

a $=$ input$(\text{'}$Enter left point of interval: $\text{'})$;

b $=$ input$(\text{'}$Enter right point of interval: $\text{'})$;

epsilon $=$ input$(\text{'}$Enter the error tolerance: $\text{'})$; $\%$ error of tolerance

x$\_$tilde $=$ a;

abs$\_$f $=$ abs$($f$($x$\_$tilde$))$; $\%$ Absolute value of f$($x$)$ at the estimated solution.

iter $=0$; $\%$ Number of iterations

fprintf$(1,$'Iteration: $\%$d $|$ Lower Interval: $\%2\mathrm{.}3$f $|$ Upper Interval: $\%2\mathrm{.}3$f $|$ Root:

$\%2\mathrm{.}3$f $|$ f$($x$)$ Abs. Value: $\%1\mathrm{.}4$f

$\text{'},...$

iter,a$,$b$,$x$\_$tilde,abs$\_$f$)$;

if$($f$($a$)*$f$($b$)>0)$

error$(\text{'}$Enter a valid interval. Abort the procedure.$\text{'})$;

else

while$($abs$\_$f $>$ epsilon$)$

x$\_$tilde$\_$next $=($a $+$ b$)/2$;

iter $=$ iter $+1$;

abs$\_$f $=$ abs$($f$($x$\_$tilde$\_$next$))$;

fprintf$(1,$'Iteration: $\%$d $|$ Lower Interval: $\%2\mathrm{.}3$f $|$ Upper Interval: $\%2\mathrm{.}3$f $|$

Root: $\%2\mathrm{.}3$f $|$ f$($x$)$ Abs. Value: $\%1\mathrm{.}4$f

$\text{'},...$

iter,a$,$b$,$x$\_$tilde$\_$next,abs$\_$f$)$;

if f$($x$\_$tilde$\_$next$)*$f$($b$)<mn>0</mn>$

a $=$ x$\_$tilde$\_$next;

else

b $=$ x$\_$tilde$\_$next;

end

x$\_$tilde $=$ x$\_$tilde$\_$next;

end

end

fprintf$(\text{'}$The root is $\%2\mathrm{.}4$f$.$

$\text{'},$x$\_$tilde$)$;Problem $1$: A Hybrid Bracketing Method $[1$ pt$]$

Review the sample code for bisection method and false position method, which can be located on Canvas

under File$/$Sample MATLAB Code. Run a few examples on your own and see if you understand the code.

If not, speak with the instructors before proceeding with this problem.

A new method for solving a nonlinear equation $f(x)=0$ is proposed in this problem. The method is a

combination of the bisection and the false position methods. The solution starts by defining an interval

$a,b$ that brackets the solution. Then estimated numerical solutions are determined once with the bisection

method and once with the false position method. $($The first iteration uses the bisection method.$)$ Write a

MATLAB user$-$defined function that solves a nonlinear equation $f(x)=0$ with this new method. Name

the function $x=$ BiFalRoot$($Fun$,a,b,$Tol $),$ where the output argument x is the solution. The input

argument Fun is a name for the function that calculates $f(x)$ for a given $x($it is a dummy name for the

function that is imported into BiFalRoot$),$ a and b are two initial points that bracket the root, and Tol is

the error tolerance.

Your code should include the following features:

The program should stop when the estimated relative error drops below the specified tolerance Tol.

Check if points a and $b$ are on opposite sides of the solution. If not, the program should halt and

display an error message of your choice.

The number of iterations should be limited to $100($to avoid an infinite loop$).$ If a solution with the

required accuracy is not obtained in $100$ iterations, the program should stop and display an error

message of your choice.

Use the function BiFalRoot to solve the equation $cos(x)-0\mathrm{.}8x^2=0$ with $a=0$ and $b=2$ and observe the

answer. For Tol use $0\mathrm{.}00001.$