# Thread: intermediate value theorem/rolle's theorem

1. ## 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].

2. 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?

3. 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 $\displaystyle x^5 - 6x + c = 0$??

4. 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 $\displaystyle g(x) = (1/6)x^6 - 3x^2 + c$ on $\displaystyle [-1,1]$ then $\displaystyle g(-1)=g(1)$ thus there is a $\displaystyle k\in (-1,1)$ such that $\displaystyle g'(k)=0 \implies f(k)=0$.

5. Originally Posted by colby2152
Is that supposed to be $\displaystyle x^5 - 6x + c = 0$??
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 $\displaystyle x^5 - 6x + c = 0$, 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 $\displaystyle [-1,1]$. 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 $\displaystyle r_1$ and $\displaystyle r_2$, with $\displaystyle r_1 < r_2$ (without loss of generality). Then we have $\displaystyle f(r_1) = f(r_2) = 0$. Thus, by Rolle's theorem, there exists an $\displaystyle r$ in the interval $\displaystyle (r_1,r_2)$ such that $\displaystyle f'(r) = 0$. Now $\displaystyle f'(x) = 5x^4 - 6$. But there is no real $\displaystyle x$ for which this attains zero in $\displaystyle [-1,1]$ much less $\displaystyle (r_1,r_2) \subset [-1,1]$. Thus we have a contradiction. Therefore, $\displaystyle x^5 - 6x + c = 0$ has at most one root in $\displaystyle [-1,1]$

QED

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

7. 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.