# show solution exists in the interval

• Jun 6th 2012, 07:07 PM
sluggerbroth
show solution exists in the interval
2x^(7)=1-x (0,1)????
• Jun 6th 2012, 07:35 PM
Reckoner
Re: show solution exists in the interval
Quote:

Originally Posted by sluggerbroth
2x^(7)=1-x (0,1)????

Are you familiar with the Intermediate Value Theorem?

We have \$\displaystyle 2x^7+x-1=0\$. Consider the function \$\displaystyle f(x)=2x^7+x-1\$. We know that \$\displaystyle f(0) = -1\$ and \$\displaystyle f(1) = 2\$. Since \$\displaystyle f\$ is continuous, what does this tell us?
• Jun 6th 2012, 07:53 PM
sluggerbroth
Re: show solution exists in the interval
i'm just learning IVT. not sure what ypu are saying about interval (0,1)
• Jun 6th 2012, 08:23 PM
Sarasij
Re: show solution exists in the interval
Quote:

Originally Posted by sluggerbroth
i'm just learning IVT. not sure what ypu are saying about interval (0,1)

What Reckoner pointed is that there is some "c" where 0 < c < 1 for which f(c) must be 0. In other words there exists atleast one root in (0,1).