# Thread: system of equations

1. ## system of equations

First off my last question was answered in such an awesome way! I am so thankful for this site- You see I'm in the hospital and am trying to complete my math homework for my class via laptop, I only have minimal calculator functions on this thing and no book to reference the how to's- so I can understand it. The hw is do monday regardless of my condition as says my teacher so this site is a heaven sent. Anyway need help with this one:

y= x^3-x
y= e^x

solve the system of equations using any method

2. i just want to make sure the exponents are correct here.

the first equation is:

$y = x^{3} - x$

right?

3. yes the first has an exponent of 3 and the second is e to the x power

4. Originally Posted by racersteve47
First off my last question was answered in such an awesome way! I am so thankful for this site- You see I'm in the hospital and am trying to complete my math homework for my class via laptop, I only have minimal calculator functions on this thing and no book to reference the how to's- so I can understand it. The hw is do monday regardless of my condition as says my teacher so this site is a heaven sent. Anyway need help with this one:

y= x^3-x
y= e^x

solve the system of equations using any method
First seeing that there are two solutions to this equation, we first note that

if $f(x)=x^3-x-e^x$

$f(2)<0$ and $f(2.5)>0$

Thus since this function is continuous we can see by the Intermediate Value Theorem that

$\exists{c}\in(2,2.5)\backepsilon~f(c)=0$

Using the Intermediate Value Theorem again but this time for the second zero we see that

$\exists{c}\in(4,4.5)\backepsilon~f(c)=0$

So now using $2.25$ and $4.25$ as intital guesses in Newton's method we see that

$\text{As }n\to\infty~~f_n(x_n)\to2.3217$

and

$f_{n_1}(x_{n_1})\to{4.3717}$ respectively

Therefore $f(4.3717)\approx{f(2.3217)}\approx{0}$