Show that the equation $1+2x+x^3+4x^5=0$ has exactly one real root

[Chris L T521]

This week's POTW is another milestone for MHB because it was two years ago that Jameson and I started this tradition on MHB. We have come quite a long ways with this and even though Jameson is no longer in charge of the Secondary level questions (anemone has done a fantastic job since taking over), the future of our POTW looks bright. Thank you again for participating in our POTWs these last two years and we look forward to your participation in future weeks! (Smile)

And now back to our regularly scheduled programming...

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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Show that the equation $1+2x+x^3+4x^5=0$ has exactly one real root.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Ackbach, chisigma, lfdahl, magneto, MarkFL, menta oscura, and Pranav. You can find two different types of solutions below.

Mark's solution: [sp]Let:

$$f(x)=4x^5+x^3+2x+1$$

Because $f$ is a function of odd degree, we know:

$$\lim_{x\to\pm\infty}f(x)=\pm\infty$$

And since, as a polynomial, $f$ is continuous over the reals, we know therefore that $f$ has at least one real root.

Now, if we observe that:

$$f'(x)=20x^4+3x^2+2=20\left(x^2+\frac{3}{40} \right)^2+\frac{151}{80}$$

We see that for all real $x$, we have:

$$0<\frac{151}{80}\le f'(x)$$

Which means that $f$ is strictly increasing over the reals, hence there is only one real root.[/sp]

magneto's solution: [sp]Let $f(x) := 4x^5 + x^3 + 2x + 1$. Since $f$ is a polynomial, it is continuous
and differentiable on $\mathbb{R}$. Observe $f(0) = 1 > 0$,
and $f(-1) = -6 < 0$. By the intermediate value theorem, there exists
$r \in [-1,0]$ where $f(r) = 0$, so we know $f$ has at least 1 real root.

Suppose there exists $r' \neq r$ where $f(r') = 0$. Without loss,
assume $r' > r$. By the mean value theorem, there exists $r < c < r'$
where $f'(c) = 0$. (If $r' < r$, we can find $r' < c < r$ where $f'(c) = 0$).
However, $f'(x) = 20x^4 + 3x^2 + 2$, which is quadratic in $x^2$.
The discriminant $\Delta = b^2 - 4ac = 9 - 4(20)(2) < 0$, so $f'(x)$ has no real roots;
hence, it is a contradiction. So, $r$ is the only real root.[/sp]