To see how scipy.optimize.fsolve() works, let's consider an example. Say you have the implicit equation: 1 x0.5 1= xe + You may spend as long as you like trying to manually solve this equation for x... or you can do this. First, cast your equation as a function of X: f(x) = xe z + 1 x0.5 def my_func (x): return x* np.exp(-x*x) + 1.0/np.sqrt (x) - 1 1 In this form, the value of x you seek, which we can tag with an asterisk, satisfies f(x) = 0 We can send this function to scipy.optimize.fsolve() and it will give us back x*: from scipy.optimize import fsolve import numpy as np. x_star = fsolve (my_func, x0=1) print ('x_star = {:.4f}; checking: f(x_star) = {:.4f).'.format (x_star [0], my_func (x_star [0]))) Here, the call to fsolve takes the function identifier as the first positional argument, and we specify an initial guess for x* using the keyword argument x0. fsolve returns a tuple, and the first element is x*. The result: x_star = 1.4612; checking: f(x_star) = -0.0000. Using the template provided, write a Python program that will find the solution to = e* - x tan(x) and print the result the same way as the example above. The template script provided below has a __main____ block that illustrates how do_implicit_solve() is to be used. To complete this lab, you are to define two functions in this script: 1. my_func(): the implicit equation cast as a function of its independent variable, which you can pass to fsolve(); and 2. do_implicit_solution (), which uses fsolve to find the solution, and which has the following specifications: One positional parameter is the identifier of the Python function you will send to fsolve; One keyword parameter using the identifier xo is the initial guess for the answer, with a default value of 1. The return value is a float representing the solution value. The __main__block provided illustrates how to use do_implicit_solve(). You can modify the __main____ block however you like. Only your my_func() and do_implicit_solve() functions will be tested.