Both Exercises are needed in Python.

I went ahead and added both the bisection and secant method functions under each exercise. I keep running across errors when I try to implement my code so please help.

EXERCISE $13\mathrm{.}2$

Define a lambda function that represents a cubic polynomial:$$y$($x$)=$x$^3-$x$-2$x

Use this lambda function to generate y values for x values ranging from $0$to $3($inclusive$)$with an increment of $0\mathrm{.}1.$

Use the bisection$\_$method function defined above to solve the root in the braket of $[0,3]$with the default stopping criteria value.

Create a line plot to visualize the cubic function.

Add appropriate labels for the x$-$axis and y$-$axis, a title for the plot, and a grid for better readability.

Mark the root in the plot.

#EXAMPLE

def bisection$\_$method$($f$,$ a$,$ b$,$ es$=1$e$-5)$:

import numpy as np

i$=0$

c$=($a$+$b$)/2$ #first guess

ea$=100$

if f$($a$)*$f$($b$)>0$:

return 'initial interval does not include solution'

#$\%\%$ iterative

while ea$>$es:

if f$($a$)*$f$($c$)>0$:

a$=$c

else:

b$=$c

i$=$i$+1$

old$=$c #save old guess

c$=($a$+$b$)/2$ #new guess

ea$=$abs$(($c$-$old$)/$c$)*100$ #relative error

print$($c$,$ea$,$i$)$

return c

print$(\text{'}$The root lies at$\text{'},$ c$)$

EXERCISE $13\mathrm{.}3$

Define a lambda function that represents a cubic polynomial:$$y$($x$)=$x$^3-$x$-2$x

Use the lambda function to generate y values for x values ranging from $0$to $3($inclusive$)$with an increment of $0\mathrm{.}1.$

Use the secant$\_$method function defined above to solve the root in the braket of $[0,3]$with the default stopping criteria value.

Create a line plot to visualize the cubic function.

Add appropriate labels for the x$-$axis and y$-$axis, a title for the plot, and a grid for better readability.

Mark the root in the plot.

#EXAMPLE

def secant$\_$method$($f$,$ x$0,$ x$1,$ es$=1$e$-6,$ maxiter$=100)$:

# Implements the Secant Method to find the root of a function f$.$

# Parameters:

# f : function, the mathematical function for which to find the root

# x$0,$ x$1$ : float, initial guesses

# es : float, tolerance for convergence $($default: $1$e$-6)$

# max$\_$iter : int, maximum number of iterations $($default: $100)$

# Returns:

# x$2$ : float, the approximated root

#initialze

ea$=100$

i$=0$

while ea$>$es and i