Show that the equationx5 − 6x + c = 0
where c is a constant has at most one real root on the interval [−1, 1].
I don't know
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