# Continuity and Roots tough question

• Oct 4th 2009, 02:08 PM
Kataangel
Continuity and Roots tough question
I've got a question to use the notion of continuity to show that $\displaystyle x^7=x-2$ has at least one real root, that is, there is at least one value of x that solves this equation.

If I try to zero one side, I'm still not sure how to get the roots"
$\displaystyle x^7-x+2=0$

I think it involves the continuity definition which I am aware of below (trying to get the x goes to a below lim)
$\displaystyle \lim{x \to a}f(x)=f(a)$

So I don't know how that will fit in with my problem above. How would I find the solution to it then?
• Oct 4th 2009, 02:17 PM
Plato
Quote:

Originally Posted by Kataangel
I've got a question to use the notion of continuity to show that $\displaystyle x^7=x-2$ has at least one real root, that is, there is at least one value of x that solves this equation.

Let $\displaystyle f(x)=x^7-x-2$.
Now $\displaystyle f$ is continuous and $\displaystyle f(1)=-3~\&~f(2)=124$
That tells us that there is root between $\displaystyle 1~\&~2$.
• Oct 4th 2009, 02:18 PM
skeeter
Quote:

Originally Posted by Kataangel
I've got a question to use the notion of continuity to show that $\displaystyle x^7=x-2$ has at least one real root, that is, there is at least one value of x that solves this equation.

If I try to zero one side, I'm still not sure how to get the roots"
$\displaystyle x^7-x-2=0$

I think it involves the continuity definition which I am aware of below (trying to get the x goes to a below lim)
$\displaystyle \lim{x \to a}f(x)=f(a)$

So I don't know how that will fit in with my problem above. How would I find the solution to it then?

let $\displaystyle f(x) = x^7 - x - 2$

note that $\displaystyle f(1) = 1 - 1 - 2 = -2 < 0$

note that $\displaystyle f(2) = 2^7 - 2 - 2 = 124 > 0$

since f(x) is continuous between x = 1 and x = 2, what function value should exist between f(1) = -2 and f(2) = 124 ?
• Oct 4th 2009, 03:37 PM
Kataangel
Sorry it seems that the actual equation is:
$\displaystyle x^7-x+2=0$

So now I've just calculated that f(-1)= 2 and f(-2)=-124.

Is there an actual way I can do this more formally than just substituting 2 close numbers?
• Oct 4th 2009, 04:42 PM
skeeter
Quote:

Originally Posted by Kataangel
Sorry it seems that the actual equation is:
$\displaystyle x^7-x+2=0$

So now I've just calculated that f(-1)= 2 and f(-2)=-124.

Is there an actual way I can do this more formally than just substituting 2 close numbers?

you are not required to find the actual root ... all you need do is prove that it exists.