In this task, a simple implementation of the bisection method for finding roots of continuous functions is to be developed. Let f: $[$a$,$ b$],$ Real numbers be a continuous function with a $<$ b and f$($a$)*$f$($b$)<0.$ According to the intermediate value theorem, f has at least one root in $[$a$,$ b$].$

The bisection method works as follows:

$1.$ Set L $=$ a$,$ R $=$ b$.$

$2.$ If the condition f$($L$)*$ f$($R$)<=0$ is not met, terminate. Otherwise:

$.$ Test whether R $-$ L $<$ TOL, where TOL is user tolerance $($e$.$g$.,$ TOL $=10^-7).$ If yes, there is a root in $[$L$,$ R$]$; otherwise,

$.$ Continue with subintervals $[$L$,($L$+$R$)/2]$ and $[($L$+$R$)/2,$ R$]$ following step $2.$

Write a recursive function

$`$void bisect$($double L$,$ double R$,$ double TOL$)`$

implementing the bisection method, and a double function $`$f$($double x$)`$ returning f$($x$).$

Test your implementation with various functions and intervals, outputting the intervals containing roots and the function values at the midpoint.

For instance, running with f$($x$)=$ x$^3-4$x$^2+6$x $-25$ and L $=0,$ R $=7$ might produce output like:

$```$

"A root of f lies in $[4\mathrm{.}044723$e$+00,4\mathrm{.}044723$e$+00]$ f$(($L$+$R$)/2)-2\mathrm{.}074738$e$-08."$

$```$

For f$($x$)=$ sin$($x$),$ you might get output similar to:

$```$

"A root of f lies in $[0\mathrm{.}000000$e$+00,5\mathrm{.}215406$e$-08]$ f$(($L$+$R$)/2)-2\mathrm{.}607703$e$-08"$

"A root of f lies in $[3\mathrm{.}141593$e$+00,3\mathrm{.}141593$e$+00]$ f$(($L$+$R$)/2)=-1\mathrm{.}741103$e$-09"$

"A root of f lies in $[6\mathrm{.}283185$e$+00,6\mathrm{.}283185$e$+00]$ f$(($L$+$R$)/2)--2\mathrm{.}259483$e$-08."$

$```$