Problem $6$: Finding a Root of a Polynomial Function

Consider the function f$($x$)=$ x$^34$x$^2+3$x $4.$ 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$^34$x$^2+3$x $4=0$ 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:

$1.$ Create variables x$1$ and x$2,$ 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.

$2.$ Create a variable named val$1,$ setting it equal to the value you would obtain by plugging x$1$ into the function f$.$

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

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

$3.$ 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.

$4.$ 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 x$1$ and x$2.$

$$ Create a variable named new$\_$val that is equal to f$($new$\_$x$).$

$$ If new$\_$val is negative, then set x$1$ to new$\_$x and set val$1$ to new$\_$val. Otherwise, set x$2$ to new$\_$x

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

$5.$ When the loop is finished executing, x$1$ and x$2$ 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 xxxx strings replaced with the appropriate values. Round the approximate solution to

seven decimal places.

The approximate solution is x $=$ xxxx$.$

The algorithm took xxxx iterations to converge.

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