# intermediate value theorem/rolle's theorem

• Dec 3rd 2007, 05:51 PM
singh1030
intermediate value theorem/rolle's theorem
Show that the equation
x
5 6x + c = 0

where
c is a constant has at most one real root on the interval [1, 1].
• Dec 3rd 2007, 05:54 PM
Jhevon
Quote:

Originally Posted by singh1030
Show that the equation
x
5 6x + c = 0

where
c is a constant has at most one real root on the interval [1, 1].

i think you need to clarify the problem. why is there an x vertically above the 5?
• Dec 3rd 2007, 06:28 PM
colby2152
Quote:

Originally Posted by Jhevon
i think you need to clarify the problem. why is there an x vertically above the 5?

Is that supposed to be $x^5 - 6x + c = 0$??
• Dec 3rd 2007, 06:57 PM
ThePerfectHacker
Quote:

Originally Posted by singh1030
Show that the equation
x
5 6x + c = 0

where
c is a constant has at most one real root on the interval [1, 1].

Let $g(x) = (1/6)x^6 - 3x^2 + c$ on $[-1,1]$ then $g(-1)=g(1)$ thus there is a $k\in (-1,1)$ such that $g'(k)=0 \implies f(k)=0$.
• Dec 3rd 2007, 07:05 PM
Jhevon
Quote:

Originally Posted by colby2152
Is that supposed to be $x^5 - 6x + c = 0$??

I don't know

Quote:

Originally Posted by singh1030
Show that the equation
x
5 6x + c = 0

where
c is a constant has at most one real root on the interval [1, 1].

well, since you refuse to respond, i will assume you meant $x^5 - 6x + c = 0$, if not. use this as a guide to do the right problem.

Proof

Assume to the contrary that there are more than one root of the equation in the interval $[-1,1]$. That is, there are at least two roots in this interval. Pick any two roots, if there are only two, pick both. Let the roots be $r_1$ and $r_2$, with $r_1 < r_2$ (without loss of generality). Then we have $f(r_1) = f(r_2) = 0$. Thus, by Rolle's theorem, there exists an $r$ in the interval $(r_1,r_2)$ such that $f'(r) = 0$. Now $f'(x) = 5x^4 - 6$. But there is no real $x$ for which this attains zero in $[-1,1]$ much less $(r_1,r_2) \subset [-1,1]$. Thus we have a contradiction. Therefore, $x^5 - 6x + c = 0$ has at most one root in $[-1,1]$

QED
• Dec 7th 2007, 07:35 AM
singh1030
to find the two inital roots by the IVT don't we need to know what the value of c is?
• Dec 8th 2007, 01:55 PM
Jhevon
Quote:

Originally Posted by singh1030
to find the two inital roots by the IVT don't we need to know what the value of c is?

no. notice that the punch line of my proof was using the derivative. it does not matter what c is here, since it would not matter in the derivative.