intermediate value theorem/rolle's theorem

**singh1030** · December 3rd 2007, 05:51 PM

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].

**Jhevon** · December 3rd 2007, 05:54 PM

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?

**colby2152** · December 3rd 2007, 06:28 PM

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 $x^5 - 6x + c = 0$ ??

**ThePerfectHacker** · December 3rd 2007, 06:57 PM

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 $g'(k)=0 \implies f(k)=0$ .

**Jhevon** · December 3rd 2007, 07:05 PM

Originally Posted by colby2152

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

I don't know

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

**singh1030** · December 7th 2007, 07:35 AM

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

**Jhevon** · December 8th 2007, 01:55 PM

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.

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