Consider the function $f(x)=x^3-4x^2+3x-4.$ A plot of this function for values of $x$ between $0$ and $5$ is provided below.

The plot above shows that the function $f(x)$ is equal to zero somewhere between $x=0$ and $x=5.$ In other words, the

plot shows that the equation $x^3-4x^2+3x-4=0$ has a solution somewhere between $x=0$ and $x=5.$ Your goal in

this problem will be to approximate the value of this solution.

Perform the following steps in a single code cell:

Create variables $x1$ and $x2,$ setting them equal to $0$ and $5,$ respectively. We will use a loop to update the values

of these variables to close in on to the solution.

Create a variable named val$1,$ setting it equal to the value you would obtain by plugging $x1$ into the function $f.$

Create a variable named val$2,$ setting it equal to the value you would obtain by plugging $x2$ into the function $f.$

Notice that we can see from the plot that val$1$ should be negative and val$2$ should be positive.

Create a variable named $n$ and set it equal to $0.$ We will use this variable to count the number of iterations

required for the algorithm to converge to a solution.

Use a loop to iteratively update the values of the four variables above as described below. This loop should

continue to execute until the absolute value of val$1$ and val$2$ are both less than $0\mathrm{.}000001.$ Each time the loop

executes, perform the following steps:

Increment $n.$

Create a variable named new$\_$x that is equal to the average of $x1$ and $x2.$

Create a variable named new$\_$val that is equal to new$\_x).$

If new$\_$val is negative, then set $x1$ to new$\_x$ and set val$1$ to new$\_$val. Otherwise, set $x2$ to new$\_$x

and set val$2$ to new$\_$val.

When the loop is finished executing, $x1$ and $x2$ should be very near each other, and near to the solution. Take

the average of the two values to be the approximation for the solution. Print your result with a message as

shown below, with the $$ strings replaced with the appropriate values. Round the approximate solution to

seven decimal places.

The approximate solution is $x=.$

The algorithm took xxxx iterations to converge.

No lists should be created for this problem, and only one loop should be used.