# Thread: solution of equation/ rolle's theorem

1. ## solution of equation/ rolle's theorem

For c > 0, prove that the following equation does not have two solutions:

$\displaystyle x^3 - 3x + c = 0$ $\displaystyle ; 0 < x < 1$
I think we have to show this using Rolle's theorem, but I am not being able to use the theorem properly on this question. Can anyone give any suggestions?

2. The LHS is decreasing (differentiate). Can a decreasing function have two zeros?

The LHS is decreasing (differentiate). Can a decreasing function have two zeros?

How is the function decreasing? Its derivative is positive, doesn't it mean that it is strictly increasing?

4. Well, if its increasing, it can't have two zeros either. Is the derivative really positive?

edit: no, I'm pretty sure its negative.

Well, if its increasing, it can't have two zeros either. Is the derivative really positive?

edit: no, I'm pretty sure its negative.
so differentiating gives $\displaystyle 3x^2 - 3 = 0$. But then, $\displaystyle f'(x) = 0$ for $\displaystyle x= \pm 1$.
So , I would do the same for $\displaystyle x^5 + 5x + 1 = 0 ; -1<x<0$ to show that the equation has only one solution..?