Question

Question #90cf3

Answer 1

To find the roots of equations like `e^x = x^3`, I recommend that you use a recursive numerical analysis method, called Newton's Method

Explanation:

Let's do an example.

To use Newton's method, you write the equation in the form `f(x) = 0`:

`e^x - x^3 = 0`

Compute `f'(x)`:

`e^x - 3x^2`

Because the method requires that we do the same computation many times, until it converges, I recommend that you use an Excel spreadsheet; the rest of my answer will contain instructions on how to do this.

Enter a good guess for x into cell A1. For this equation, I will enter 2.

Enter the following into cell A2:

=A1-(EXP(A1) - A1^3)/(EXP(A1) - 3*A1^2)

Please notice that the above is Excel spreadsheet language for

`x_2 = x_1 - (e^(x_1)-x_1^3)/(e^(x_1)-3x_1^2)`

Copy the contents of cell A2 into A3 through A10. After only 3 or 4 recursions, you can see that the method has converged on

`x = 1.857184`

Answer 2

We can use the Intermediate Value Theorem to see that each pair has at least one point of intersection.

Explanation:

`f(x) = e^x-x^2` is continuous on the entire real line.

At `x=0`, we have `f(0)=1`.

At `x=-1`, we have `f(-1) = 1/e-1` which is negative.

`f` is continuous on `[-1,0]`, so there is at least one `c` in `(-1,0)` with `f(c)=0`.

`g(x)=e^x-x^3` is continuous on the entire real line.

At `x=0`, we have `g(0)=1`.

At `x=2`, we have `g(2) = e^2-8` which is negative.
(Note that `e^2 ~~ 2.7^2 < 7.3 < 8`.)

`g` is continuous on `[0,2]`, so there is at least one `c` in `(0,2)` with `g(c)=0`.