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.
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.