1. ## derivations.

Prove that the equation x^3-3x+c=0 can't have 2 different soulutions in the segment [0,1].
It is suggested to find the derivation and to use the way of supposing
from the contrary.

Any help is appreciated and sorry for the language.

2. The derivative of
$\displaystyle \frac{d}{dx} (x^3 - 3x + C ) = 3x^2 - 3$

which is always negative in the interval [0,1).

So if the function went through the y=0 line in that interval once, it can't turn around and go back through it to make a second root.

3. Originally Posted by blertta
Prove that the equation x^3-3x+c=0 can't have 2 different soulutions in the segment [0,1].
It is suggested to find the derivation and to use the way of supposing
from the contrary.

Any help is appreciated and sorry for the language.
Intermediate value theorem

4. Originally Posted by qmech
The derivative of
$\displaystyle \frac{d}{dx} (x^3 - 3x + C ) = 3x^2 - 3$

which is always negative in the interval [0,1).

So if the function went through the y=0 line in that interval once, it can't turn around and go back through it to make a second root.
i think the function is strictly monotone on [1,0),so there's a unique solution
$\displaystyle b$ such that $\displaystyle b\in [2+c,-2+c]$.
no ?