Question #90cf3

2 Answers
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(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#

Sep 26, 2016

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#.