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Suppose $p$ is a polynomial with real coefficients. Then which of the following statements are necessarily true?

  1. There is no root of the derivative $p$' between two real roots of the polynomial $p$.

  2. There is exactly one root of the derivative $p$' between any two real roots of the polynomial $p$.

  3. There is exactly one root of the derivative $p$' between any two consecutive real roots of the polynomial $p$.

  4. There is at least one root of the derivative $p$' between any two consecutive roots of $p$

I have taken $p(x)=x^2-1$ then $p'(x)=2x$, here 0 is the root of $p'$ that is lying between the roots of $p$ that is -1 and 1. Hence option 1 is wrong. For option 2, i have chosen $p(x)=x$. I guess that option 3 is true but i am not able to prove. Also i am not able to give example for option 4. Please help me!

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    $\begingroup$ Note that taking examples is used to prove contradictions and not for proving a necessarily true statement. $\endgroup$
    – Arpan
    Apr 23, 2015 at 10:09
  • $\begingroup$ i am just omiting the options. $\endgroup$ Apr 23, 2015 at 10:12
  • $\begingroup$ Here using the fundamental theorem of algebra could simplify it are you allowed to use it $\endgroup$
    – happymath
    Apr 23, 2015 at 10:12
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    $\begingroup$ For your option 2, you chose $P(x)=x$, but it does not have two real roots, so you cannot omit it in that way. (Although it is incorrect). $\endgroup$
    – Arpan
    Apr 23, 2015 at 10:13
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    $\begingroup$ $x^4 - 3x^2 -4 = (x-2)(x+2)(x^2 + 1)$ is a counterexample to the first three points simultaneously. $\endgroup$
    – Arthur
    Apr 23, 2015 at 10:15

1 Answer 1

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Hint: Rolle's theortem

If a real-valued function $f$ is continuous on a proper closed interval $[a, b]$, differentiable on the open interval (a, b), and $f(a) = f(b)(=0)$, then there exists at least one c in the open interval $(a, b)$ such that $f'(c) = 0$.

Now, what about a polynomial and two consequtive roots of it? (The first three statements can be falsified by simple examples.)

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  • $\begingroup$ I was just adding this as a hint in the comments too. Guess I was too slow. $\endgroup$
    – HSN
    Apr 23, 2015 at 10:22
  • $\begingroup$ Thank u zoli. Now i got the point. $\endgroup$ Apr 23, 2015 at 10:23
  • $\begingroup$ @HSN, Sorry! My answer may not deserve the "answer" status. But I was getting bored with the already existing comments. $\endgroup$
    – zoli
    Apr 23, 2015 at 10:24

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