Solve in MATLAB using the starter code and function to get the provided output. The specific growth rate of a yeast that produces antibiotic is a function of the food concentration $c$ :

$g=\frac{2c}{4+0\mathrm{.}8c+c^2+0\mathrm{.}2c^3}$

As depicted in the following figure:

growth goes to zero at very low concentrations due to food limitation$.$ It also goes to zero at high concentrations due to toxicity effects. Find that value of $c$ at which growth is a maximum using MinParabolicinterpolation.mlx$,$ and plot the results.

xopt $=1\mathrm{.}5679$

fopt $=0\mathrm{.}3696$

function $[$xopt$,$ fopt$]=$MinParabolicInterpolation $($func$,$ xlow, xhigh, es$,$ maxit, varargin$)$

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

if nargin $<mn\; id="EditorElement\_1640680">4</mn>$ l lisempty$($es$),$ es$=0\mathrm{.}0001$; end

if nargin$<mn\; id="EditorElement\_1640636">5</mn>$I lisempty $($maxit$),$ maxit$=50$;end

iter $=0$;

$x1=$xlow;$x3=$xhigh;

$x2=\frac{x1+x3}{2}$;

$,$ varargin $\{\begin{array}{c}?\end{array}$

$f2=$ func $,$ varargin $\{\begin{array}{c}?\end{array}$;

$f3=$ func $,$ varargin$\{$;

if $f2<mn\; id="EditorElement\_1640343">1</mn>$ & & xoptold $=x2$; end

while $(1)$

xopt$=x2-{0\mathrm{.}5}^{***}\frac{(x2-x1)2^{***}(f2-f3)-(x2-x3{)}^{{?}^{{?}^{?}}}{2}^{***}(f2-f1)}{(x2-x1{)}^{***}(f2-f3)-(x2-x3{)}^{***}(f2-f1)}$;

fopt $=$ func$($xopt$,$ varargin $\{$;

iter $=$ iter $+1$ ;

if $x$ opt $>x2$

$x1=x2$;

$f1=f2$;

else

$x3=x2$;

$f3=f2$;

end

x$2=$ xopt; $2=$ fopt;

if xopt $=0,$ ea$=$abs $(($xopt $-$ xoptold$)/$ xopt$)*100$ ;end

xoptold $=$ xopt;

if eaes II iter$>=$maxit,break, end

end

$1$

end