prove that x^3-3x+1=0 has exactly two positive roots and one negative root.
Please don't tell me look at the graph,because I've already seen it and I need to prove it(which means that the graph says nothing to me).
+118725 x^3-3x+1=0
Let y = x^3-3x+1
y'=3x^2-3
y''=6x
Find turning points y'=0
3x^2-3=0
3(x^2-1)=0
3(x-1)(x+1)=0
x=1 and x=-1
When x=1 y=1-3+1=-1
When x=-1 y=-1+3+1=3
So there is 1 root between 1 and -1
When x=0 y=0-0+1=1
So there is 1 root between 0 and 1
Consider concavity
y''=6x
If x>0 y''>0 concave up
thereforethere will be another root that is greater than 1
If x<0 y'' < 0 concave down
Therefore there will be another root that is less than -1
There can only be a maximum of 3 roots because concavity onle chatnges once.
So there are 2 positive roots and 1 negative root.
I am not sure if this is what you are expected to produce but it works for me. 
+118725 x^3-3x+1=0
Let y = x^3-3x+1
y'=3x^2-3
y''=6x
Find turning points y'=0
3x^2-3=0
3(x^2-1)=0
3(x-1)(x+1)=0
x=1 and x=-1
When x=1 y=1-3+1=-1
When x=-1 y=-1+3+1=3
So there is 1 root between 1 and -1
When x=0 y=0-0+1=1
So there is 1 root between 0 and 1
Consider concavity
y''=6x
If x>0 y''>0 concave up
thereforethere will be another root that is greater than 1
If x<0 y'' < 0 concave down
Therefore there will be another root that is less than -1
There can only be a maximum of 3 roots because concavity onle chatnges once.
So there are 2 positive roots and 1 negative root.
I am not sure if this is what you are expected to produce but it works for me.
+33666 As a matter of interest why shouldn't a graph be used as a proof? Given that a cubic polynomial has exactly three roots it is blindingly obvious from the graph of this function that it has one negative root and two positive roots!
.
Descartes' rule of signs gets you there, if you accept it as a proof that is ?
\(p(x)=x^{3}-3x+1\)
has two sign changes, + to -, then - to + so \(p(x)=0\) has 2 or 0 positive roots.
\(p(-x)=-x^{3}+3x+1\)
has one sign change so \(p(x)=0\) has 1 negative root.
It follows that the equation has two positive roots and one negative root.
Well,the exercise needs a solution that has to do with bolzano's equations or with continuous functions or the equations that are the aftermath of bolzano's theories,such as m<=f(Xo)<=M
I think that for the Descarts' 'proof ' , the zero number of positive roots has to be ruled out.
\(p(0)=1, p(1)=-1, p(2) =3\)
is sufficient to show that.
+130555 Using Melody's answer as a guide, the function has a max at (-1, 3)
And since the end behavior of the function as it approaches infinity is negative.......it has to have at least one negative root, because it would have to cross the x axis once as it approaches x = -1 from the left side
The y intercept is positive......and we have a min at (1, -1)......so the function has to cross the x axis once between x = 0 and x = 1
Likewise, the end behavior of the function is positive as is approchaes positive infinity......so....it must cross the x axis somewhere between x = 1 and infinity
Thus......the function has one negative root and two positive ones........
+118725 Well,the exercise needs a solution that has to do with bolzano's equations or with continuous functions or the equations that are the aftermath of bolzano's theories,such as m<=f(Xo)<=M
If you knew that is what you needed why didn't you say so. !!
I/We did not know whether you were year 9 or whether you were in University.
When you said you did not want a graph I thought you meant that you did not want a computer generated graph that you did not understand.
Please express your questions better!
