Solving for Real Roots of a Polynomial Equation: Using the Mean Value Theorem

[ace123]

[SOLVED] Mean value theorem

First I just want to say that my professor hasn't gotten up to teaching us this so I may be a little slow in understanding this material and want to thank you for being patient with me.

The question asks to show that the equation X^4 -4X + c = 0 has at most two real roots.

Now I believe that c is just a constant and I just applied the Mean value theorem and got c= \sqrt[3]{2}

So I believe that's one of my roots because I chose a= 0 and b= 2 and it lies in the interval. But now I'am at a lose as to how to find a 2nd root. All my examples in the textbook show how to find 1 real root.

I' am also not sure if what I have done so far is entirely correct. Thanks for any help, I just want a hint as to how to find a second root

Edit: can someone delete the other one I don't know why their are 2 of them I guess i clicked twice, sorry..

 

[HallsofIvy]

First I just want to say that my professor hasn't gotten up to teaching us this so I may be a little slow in understanding this material and want to thank you for being patient with me.

The question asks to show that the equation X^4 -4X + c = 0 has at most two real roots.

Now I believe that c is just a constant and I just applied the Mean value theorem and got c= \sqrt[3]{2}

What? You are asked to that x⁴- 4x+ c= 0 has at most two real roots no matter what c is. I don't know what you mean by "got c= \sqrt[3]{2}! Did you accidently use c as a value of x when it was already in the equation?

So I believe that's one of my roots because I chose a= 0 and b= 2 and it lies in the interval. But now I'am at a lose as to how to find a 2nd root. All my examples in the textbook show how to find 1 real root.

I' am also not sure if what I have done so far is entirely correct. Thanks for any help, I just want a hint as to how to find a second root

Edit: can someone delete the other one I don't know why their are 2 of them I guess i clicked twice, sorry..

You say you applied the mean value theorem but you don't say what function you applied it to! Do you mean that you found x such that f'(x)= 4x³- 4= (f(2)-f(1))/1= 16- 8+ c- (1- 4+ c)= 11? That would be solving 4x³= 15 which does NOT give x= \sqrt[3]{2}. And why did you pick a= 0, b= 2?

In any case, the problem does not ask you to find two roots of the equation- it asks you to show that the equation cannot have more than 2 real roots. It may, in fact, have none!

Suppose that x₁, x₂, and x₃ are 3 real solutions. Apply the mean value theorem to the function f(x)= x⁴- 4x+ c on the interval [x₁,x₂]. That tells you that there must be a value of x in that interval so that f'(x)= 4x³- 4= what? Now apply the mean value theorem to that function on the interval [x₂,x₃]. That tells you that there must be a value of x in that interval so that f'(x)= 4x³- 4= what? Do you see why those two statement cannot both be true?

 

[ace123]

ooo So I don't need to find the actual roots just show that their is two roots. I did the whole thing with a=0 and b= 2 because that's what the book did and I just copied the book

 

[HallsofIvy]

The book solved THAT problem for you? You surely don't expect to be able to copy the solution to some other problem and have the same solution work for yours?

 

[ace123]

I believe both statements cannot be true because then that would mean their is a number c in [x1,x2] and in the interval [x2,x3]. Which would be impossible. But I'm still a little hazy how this shows that their cannot be more than 2 roots i thought that this shows that their is not more than one because you can't have the same number in 2 different intervals

 

[ace123]

LOL. The problem was different but i assumed i could use the same concept because out of nowhere they say use a=0 and b=2 and then apply the mean value theorem to a function

 

[ace123]

ahh after reading your posts repeatedly I finally understand that if their are three answers then that would mean that the number c is in 2 different intervals which can't be.
Anyway Thanks HallsofIvy

 

[rrren]

Hi HallsofIvy,

I am going through the exact same probem and still don't understand your solution. Would you mind going through step by step how to solve it and show that it has at most two real roots? What are the X1, X2, X3 I would choose?

Thanks

 

[alkashi]

I believe both statements cannot be true because then that would mean their is a number c in [x1,x2] and in the interval [x2,x3]. Which would be impossible. But I'm still a little hazy how this shows that their cannot be more than 2 roots i thought that this shows that their is not more than one because you can't have the same number in 2 different intervals

u right we can't have the same number in different intervals. but the question is are they same? i don't think so. in fact i don't see any contradiction. applyin the mean value theorem in [x1 x2] we will a number and in [x2 x3] another wone. try to look at it deeply u will see that they r not indentic. i think that u think it s the same number because u take the number as x. look at it

 

[alkashi]

Hi HallsofIvy,

I am going through the exact same probem and still don't understand your solution. Would you mind going through step by step how to solve it and show that it has at most two real roots? What are the X1, X2, X3 I would choose?

Thanks

hey i don't know if u got it or not but this is what i do.

we are ask to show that the (E) has at most 2 real roots right? in other word is to show that we can't have 3 roots. in fact we won't even have a root for some values of c in the equation. so if i can just show that if we can't hav 3 solutions am done.

NB: just keep in mind that we are alwayz seeking a contradiction to conclude. ok now:

let a,b and d be 3 solutions of the equation. so f(a)=f(b)=f(c)=0
the fuction is a polynomial so continuous and differentiable on R . so by the Rolle's theorem ( we pratically use this theorem for all those demonstrations) there is a X1 in (a b) and a X2 in (b c) such as f '(X1)=0=f '(X2).

read very well that paragraphs. it says that if we have 3 solutions for f , we must have 2 solutions f ' (prime). ( those X1 and X2 ). ok now what do we have to do? just verify that f prime has two solutions.

f '(x)=0 ==> 4X^3 + 4=0
X^3= -1
X= -1

Whattttt? we know that for 3 solutions of f we must have 2 solutions of f prime. THERE IS A CONTRADICTIONNNNN.

i let u the pleasure to conclude.