Rolle's Theorem $x^3 - 3x +b$ Use Rolle's Theorem to prove that the equation $x^3 - 3x + b = 0$ has at most one root in the interval $[-1,1]$.

Rolle's Theorem : Suppose f is a continuous real-valued function on $[a,b]$ with $f(a) = f(b)$, and that f is differentiable on $(a,b)$. Then there exists c in $(a,b)$ such that $f'(c) = 0$.

Here is my proof:

Suppose there are 2 roots in $[-1,1]$. $f'(x) = 3(x^2-1)$, which is equal to zero at the endpoints -1 and 1. This results in a contradiction because there are are no roots in the interval $(-1,1)$. Therefore the original hypothesis that there are two roots in $[-1,1]$ is false, implying that there is at most 1 root.

I know that is incorrect because I never used the hypothesis, but I think its the right idea. Help would be more than appreciated.

  • $\begingroup$ Are you using $a=-1$ and $b=1$ ? $\endgroup$ – imranfat Nov 7 '13 at 22:06
  • $\begingroup$ There is no contradiction in what you said. f' can be 0 at -1 and 1 and f can still have roots in that interval. $\endgroup$ – Betty Mock Nov 7 '13 at 22:08
  • $\begingroup$ user105631, please don't open a second question on the same topic. If you wanted to clarify your last question it would be better to edit that one. $\endgroup$ – Antonio Vargas Nov 7 '13 at 23:09

You almost got it. The hypothesis comes from the fact that you have two roots.

Suppose for the sake of contradiction that the function $f(x) = x^3 - 3x + b$ has two (or more) roots in $[-1,1]$. Let (any two of) them be denoted as $x_1$ and $x_2$.

This means $f$ is a continuous function on $[x_1,x_2]$ with $f(x_1)=f(x_2)=0$. Rolle's theorem then implies the existence of $c\in(x_1,x_2)\subseteq(-1,1)$ such that $f'(c) = 0$. But you've shown that there is no such $c$. That is your contradiction.


Use a proof by contradiction.

Suppose that $f$ has $2$ distinct roots in $[-1, 1]$. Call them $a$ and $b$, with $a < b$. Since they are roots, $$ f(a) = 0 = f(b). $$

By Roll's Theorem, there is a real number $c \in (a, b)$ (i.e., $a < c < b$) where $f'(c) = 0$. Now look at the derivative of $f$ to arrive at a contradiction. $$ f'(x) = 3x^2 - 3 = 3(x^2-1) $$ so $f'(x)$ only vanishes at $x = \pm 1$.


Suppose -1 < q < r < 1 and f(q) = f(r) = 0. Then there is a point -1 < s < 1 such that f'(s) = 0 (Rolle's theorem). However, you have pointed out that the only places f' is 0 are at -1 and 1. Therefore there is no such s, and therefore there cannot be 2 roots.


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