# Question #90cf3

Sep 26, 2016

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 $f \left(x\right) = 0$:

${e}^{x} - {x}^{3} = 0$

Compute $f ' \left(x\right)$:

${e}^{x} - 3 {x}^{2}$

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)

${x}_{2} = {x}_{1} - \frac{{e}^{{x}_{1}} - {x}_{1}^{3}}{{e}^{{x}_{1}} - 3 {x}_{1}^{2}}$

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

$x = 1.857184$

Sep 26, 2016

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

#### Explanation:

$f \left(x\right) = {e}^{x} - {x}^{2}$ is continuous on the entire real line.

At $x = 0$, we have $f \left(0\right) = 1$.

At $x = - 1$, we have $f \left(- 1\right) = \frac{1}{e} - 1$ which is negative.

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

$g \left(x\right) = {e}^{x} - {x}^{3}$ is continuous on the entire real line.

At $x = 0$, we have $g \left(0\right) = 1$.

At $x = 2$, we have $g \left(2\right) = {e}^{2} - 8$ which is negative.
(Note that ${e}^{2} \approx {2.7}^{2} < 7.3 < 8$.)

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