f $=$ @$($x$)$ x$^3-10*$x $-5$;

xl $=2$;

xu $=4$;

fxl $=$ f$($xl$)$;

fxu $=$ f$($xu$)$;

if fxl $*$ fxu $>0$

error$(\text{'}$Initial guesses do not bracket the root.$\text{'})$;

end

$\%$ Set tolerance and maximum number of iterations

tol $=1$e$-8$;

maxIter $=1000$;

$\%$ Initialize variables

xr $=0$;

error $=$ inf;

iter $=0$;

while abs$($error$)>$ tol && iter $$ maxIter

xr$\_$prev $=$ xr;

xr $=$ xu $-$ fxu $*($xl $-$ xu$)/($fxl $-$ fxu$)$;

$\%$ Evaluate the function at xr

fxr $=$ f$($xr$)$;

if fxr $*$ fxl $<mtext\; id="EditorElement\_9458167"></mtext><mn\; id="EditorElement\_9458168">0</mn>$

xu $=$ xr;

fxu $=$ fxr;

elseif fxr $*$ fxu $<mtext\; id="EditorElement\_9458228"></mtext><mn\; id="EditorElement\_9458229">0</mn>$

xl $=$ xr;

fxl $=$ fxr;

else

break;

end

if xr ~$=0$

error $=($xr $-$ xr$\_$prev$)/$ xr;

end

iter $=$ iter $+1$;

end

$\%$ Display results

fprintf$(\text{'}$Root found at xr $=\%\mathrm{.}8$f

$\text{'},$ xr$)$;

fprintf$(\text{'}$Number of iterations: $\%$d

$\text{'},$ iter$)$;

fprintf$(\text{'}$Error: $\%\mathrm{.}8$e

$\text{'},$ abs$($error$))$;

A$5-3.$ Modify the code used in Example $4$ to find the root only at $f(x)<mn\; id="EditorElement\_9457680">0</mn><mn\; id="EditorElement\_9457681">.</mn><mn\; id="EditorElement\_9457682">0</mn><mn\; id="EditorElement\_9457683">1</mn>$ using

Newton$-$Rephson Method without showing any iteration. Also find the root of

equation, $f(x)=x^5-3x-10,$ take initial guess, $x_0=2$