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It looks to me like I finished the proof, but I want to make sure; the part near the end seems like it could be better.

The problem states: Prove that the polynomial $f(x) = x^5 + x^3 − 1$ has exactly one real root.

Polynomials are continuous and differentiable everywhere, so the Intermediate Value Theorem and Rolle's Theorem apply. Slightly arbitrarily, $f(0)=-1$ and $f(1)=1$. By the IVT, $f(a)=0$ for some $a\epsilon[0,1]$. Thus there is at least one real root.

Now, I think I'm supposed to reach a contradiction using Rolle's Theorem, so I assume $f$ has two real roots, i.e., $f(x_1)=0$ and $f(x_2)=0$. By the theorem, there is $a\epsilon(x_1,x_2)$ such that $f'(a)=0$.

$f'(x)=5x^4+3x^2$, and $f'(x)=0$ iff $x=0$. Thus $a=0$ and $0\epsilon(x_1,x_2)$. Thus $x_1<0$ and $x_2>0$.

But $x_1$ is supposed to be a root of $f(x)$; there cannot possibly be a negative root of the given polynomial. Thus the assumption that there are two is false.

Is this sufficient? At first I thought so, but I've only shown there can't be two. Couldn't there be three roots?

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    $\begingroup$ works fine if you replace 'has two roots' with 'has at least two different roots, say $x_1$ and $x_2$' $\endgroup$
    – mm-aops
    Jun 6, 2014 at 19:08

3 Answers 3

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$f^{\prime}$ is always positive so $f$ is increasing everywhere. Thus at most one root, and since it goes to plus and minus infinity there must be at least one root.

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    $\begingroup$ positive a.e. actually, still works tho $\endgroup$
    – mm-aops
    Jun 6, 2014 at 19:09
  • $\begingroup$ It has an inflection at $0$. $\endgroup$ Jun 6, 2014 at 19:11
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We have $0 \leq f'(x)$ for all real $x$, so $f$ is monotonic, so has only one root.

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  • $\begingroup$ Of course, your version seems easier. Oddly, the hint for the problem recommends using Rolle's theorem for the second part, but I don't think this method requires this, does it? $\endgroup$ Jun 6, 2014 at 19:11
  • $\begingroup$ you need it to be strictly monotonic, so the answer isn't complete, need to say that $f'=0$ at a single point doesn't spoil things. also: the fact that $f' \geq 0$ implies $f$ non-decreasing is usually deduced from Rolles / Lagrange theorem, so in a way it still uses it $\endgroup$
    – mm-aops
    Jun 6, 2014 at 19:16
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You are helped by the fact that the roots of the derivative are easily computable, so that you can decompose the domain into intervals between the extrema.

An interval contains a root iff the extrema have opposite signs.

[For complete rigor, you need to discard multiple roots by taking the gcd of $f$ and $f'$.]

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