Identify the limitations of the algorithm above.

B$.$ Study the function code given and answer the questions that follow.

function $[$root$,fx,$ ea$,$ iter$]=$bisect $($func$,$ xl$,$ xu$,$ es$,$ maxit, varargin$)$

if nargin $3,$ error$(\text{'}$at least $3$ input arguments required'$),$ end

test $=$ func $(x1,$varargin$\{\begin{array}{c}?\end{array})**$ func $(xu,$varargin$\{\begin{array}{c}?\end{array})$;

if test $>0,$ error$(\text{'}$no sign change'$),$ end

if nargin$<mn\; id="EditorElement\_40282">4</mn><mi\; id="EditorElement\_40283">|</mi><mn\; id="EditorElement\_40284">1</mn>$isempty$($es$),$ es$=0\mathrm{.}0001$; end

if nargin$<mn\; id="EditorElement\_40323">5</mn><mi\; id="EditorElement\_40324">|</mi><mi\; id="EditorElement\_40325">|</mi>$isempty $($maxit$),$ maxit$=50$; end

iter $=0$;$xr=xl$; ea $=100$;

while $(1)$

xrold$=xr$;

$xr=\frac{x1+xu}{2}$;

iter $=$ iter $+1$ ;

if $xr=0,$ ea $\frac{?}{a}=bs(\frac{xr-\text{x}\text{r}\text{o}\text{l}\text{d}}{x}r)**100$; end

test $=$ func $(x),$varargin$\{\begin{array}{c}?\end{array}$ func $(xr,$varargin$\{\begin{array}{c}?\end{array})$;

if test $<mn\; id="EditorElement\_39771">0</mn>$

$xu=xr$;

elseif test $>0$

$xl=xr$;

else

ea $=0$;

end

if ea es $||$ iter $>=$ maxit, break, end

end

root $=xr$;$fx=$func$(xr,$varargin$\{\begin{array}{c}?\end{array})$;

What does this function?

Look up the MATLAB function vararg in employed in the function script above. What is varargin and what is it used for?

What are the default values for es and maxit in the function call below?

$>$val$=$bisect$($ func $,xl,xu)$

If line $14$ is true, which interval is represented by line $15?$

if line $16$ is true, which interval odds represented by line $17?$

In line $10$ what does variabale xr represent?

In the functionn script above, what are the conditions for the loop termination? Explain