Create a user defined function called bisection $(fn,x_{-}l,x_{-}u,$ TOL, MAX$\_$ITER $)$ that implements the Bisection Method for computing the roots of any given equation.

The equation that is to be solved can be defined as an anonymous function called $)$ and this function can be used within the bisection function.

You will use your bisection function within a MATLAB script given below to find a root of an equation defined as an anonymous function. The arguments for the bisection function should be the initial guesses for the lower bound $x_{-}l$ and the upper bound $x_{-}u,$ the tolerance TOL for the aboslute error. Your bisection function should stop iterating on the root when the error TOL is less than $1{10}^{-5}$ and return the value of the root. If the convergence is not reached at the end of maximum iterations MAX$\_$ITER, set the variable convergence to false. See below for more information about inputs$/$outputs:

The input should be:

$fn$ : The function for which we want to find the root;

x$\_$l: The lower bound of the interval in which we want to search for the root;

x$\_$u: The upper bound of the interval in which we want to search for the root;

TOL: The desired tolerance for the root, which is $1{10}^{-5}$;

MAX$\_$ITER: The maximum number of iterations that the method will perform before terminating.

The returned values from the bisection function should be:

x: The approximate value of the root of the function.

iterations: The number of iterations performed by the bisection method.

convergence: A boolean variable that indicates whether the method converged to a root within the given tolerance.

When testing $($or debugging$)$ your program, it is important to solve a relatively simple problem for which you know the answer. For example, use the simple polynomial for the accelerating car $($this is already implemented below in the script$)$:

$f(t)=0\mathrm{.}5*a*t^2+v_0*t-d$

For $a=0\mathrm{.}6\frac{m}{s^2},v_0=4\mathrm{.}5\frac{m}{s},$ and $d=50m,$ the approximate solution for the root is at $t$~~$7\mathrm{.}43s.$Create a user defined function called bisection $(fn,x_{-}l,x_{-}u,$ TOL, MAX$\_$ITER $)$ that implements the Bisection Method for computing the roots of any given equation.

The equation that is to be solved can be defined as an anonymous function called $)$ and this function can be used within the bisection function.

You will use your bisection function within a MATLAB script given below to find a root of an equation defined as an anonymous function. The arguments for the bisection function should be the initial guesses for the lower bound $x_{-}l$ and the upper bound $x_{-}u,$ the tolerance TOL for the aboslute error. Your bisection function should stop iterating on the root when the error TOL is less than $1{10}^{-5}$ and return the value of the root. If the convergence is not reached at the end of maximum iterations MAX$\_$ITER, set the variable convergence to false. See below for more information about inputs$/$outputs:

The input should be:

$fn$ : The function for which we want to find the root;

x$\_$l: The lower bound of the interval in which we want to search for the root;

x$\_$u: The upper bound of the interval in which we want to search for the root;

TOL: The desired tolerance for the root, which is $1{10}^{-5}$;

MAX$\_$ITER: The maximum number of iterations that the method will perform before terminating.

The returned values from the bisection function should be:

x: The approximate value of the root of the function.

iterations: The number of iterations performed by the bisection method.

convergence: A boolean variable that indicates whether the method converged to a root within the given tolerance.

When testing $($or debugging$)$ your program, it is important to solve a relatively simple problem for which you know the answer. For example, use the simple polynomial for the accelerating car $($this is already implemented below in the script$)$:

$f(t)=0\mathrm{.}5*a*t^2+v_0*t-d$

For $a=0\mathrm{.}6\frac{m}{s^2},v_0=4\mathrm{.}5\frac{m}{s},$ and $d=50m,$ the approximate solution for the root is at $t$~~$7\mathrm{.}43s.$