Question

Question #adfe5

Answer 1

You cannot find the roots using algebraic methods. You can find the real roots using Newton's Method .

Explanation:

The Newton's Method reference gives the following equation:

`x_(n+1)=x_n-f(x_n)/(f'(x_n))" [1]"`

You must write the given function in the form `f(x_n)=0`

`f(x_n) = e^(x_n+1)-x_n-3=0" [2]"`

Compute the derivative:

`f'(x_n) = e^(x_n+1)-1" [3]"`

Enter the expressions of [2] and [3] into equation [1]

`x_(n+1)=x_n-(e^(x_n+1)-x_n-3)/(e^(x_n+1)-1)" [4]"`

Because this is a recursive operation. I use an Excel spreadsheet. Also, you can use Matlab or any other computation engine.

The following are instructions on how to do the recursion, using Excel.

It looks like there is a root between 0 and 1, therefore, we enter 0 into Cell A1. This will be our `x_0`

Enter the following into Cell A2:

=A1-(EXP(A1+1)-A1-3)/(EXP(A1+1)-1)

Please observe that this is the Excel language equivalent of equation [4].

Copy the contents of Cell A2 into cells A3 through A10.

Please observe that, after 3 recursions, the values displayed in the cells converge upon `0.146193`.

It looks like there is another root about about `x =-2` so enter -2 into Cell A1.

After 3 recursions, the values displayed in the cells converge upon `-2.84141`.

Therefore, the two real values that solve this equation are `x ~~0.146193` and `x~~ -2.84141`..

If you require more accuracy, you can expand the width of the cells .