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It's clear that this function has a zero in the interval $[-2,-1]$ by the Intermediate Value Theorem. I have graphed this function, and it's easy to see that it only has one real root. But, this function is not injective and I'm having a very hard time proving that it has exactly one real zero. I can't calculate the other 4 complex roots, and my algebra is relatively weak. I have also looked at similar questions, where the solutions use Rolle's Theorem, but I can't seem to apply it to this problem.

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  • $\begingroup$ Not sure but if f'(x)=10x^4+12x^3+2 is positive positive or negative on the interval it can't have more than one root. $\endgroup$ – fleablood Jan 7 '18 at 2:40
  • $\begingroup$ The derivative, $f'$ has two real zeros, both in the interval $[-1,0]$. Although, I can't prove these values are unique either. $\endgroup$ – Caleb Nastasi Jan 7 '18 at 2:47
  • $\begingroup$ If you know Descartes' rule of signs, it is clear that this polynomial has either three or one negative real root(s). Now we need to show that there can not be three (negative) real roots. $\endgroup$ – Bumblebee Jan 7 '18 at 3:11
  • $\begingroup$ Right. If f(x) = 0 and f(y) = 0 then f'(k) = 0 for some x < k < y. Since f'(x) = 0 means -1 < x < 0 that mus f(x) can only have at most one zero in x < -1. So [-2,-1] has only one root. Now you have to show there are no zeros for f(x) > -1 whic it clearly can't as the $2x^5 +3x^4 +2x + 16 > -2 +0 -2 + 16> 0$. $\endgroup$ – fleablood Jan 7 '18 at 3:40
  • $\begingroup$ @fleablood Nice observation. But how do we know that $f^{\prime}$ has only roots in $[-1,0]$ ? $\endgroup$ – Rene Schipperus Jan 7 '18 at 3:48
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Any real roots must be in $\,(-\infty, -1)\,$, because:

  • there can be no positive roots $\,x \ge 0\,$ since all coefficients are positive;

  • furthermore, there can be no roots with magnitude $\,1\,$ or smaller $\,x = a \in [-1,1]\,$, since $\,f(a)=2a^5+3a^4+2a+16 \ge -2+0-2+16 = 12 \gt 0\,$.

Let $\,x = -(y+1) \,$, so that $\,x \lt -1 \iff y \gt 0\,$. Substituting back:

$$\,-2(y+1)^5+3(y+1)^4-2(y+1)+16 \;=\; -2 y^5 - 7 y^4 - 8 y^3 - 2 y^2 + 15\,$$

The latter can only have one real positive root $\,y \gt 0\,$ by Descartes' rule of signs, so there is only one real root $\,x \lt -1\,$.

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    $\begingroup$ Nicely done +1! $\endgroup$ – Macavity Jan 7 '18 at 6:15
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$f(x) = 2x^5+3x^4+2x+16$ clearly cannot have non-negative roots, so let us investigate negative roots, considering $f(-x) = -2x^5+3x^4-2x+16$. It has three sign changes, so by Descartes' rule of signs this can have either $1$ or $3$ negative roots.

Then again, $f(-x) = x^4(-2x+3)+(-2x+3) + 13 = (x^4+1)(-2x+3)+13$. As $x$ increases, the only term which can cause a sign change is $-2x+3$, which can only change signs once. Hence there is only one negative root.

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we have $p(x) = 2x^5+3x^4+2x+16 = (x^4+1)(2x+3)+13=0$. So we are intersecting $f(x) = x^4+1$ with $g(x) = \frac{-13}{2x+3}$.

Obviously this two colide just once at the given interval $[-2,-1]$ that you mentioned earlier and this root is unique because when $x>0$ we have always $x^4+1 > \frac{-13}{2x+3}$ (LHS is always positive while RHS is always negative). when $x<0$ LHS is strictly decreasing and RHS is strictly increasing and knowing the fact that for large $x$ we have $x^4+1 > \frac{-13}{2x+3}$ and for $x$ close to $-1.5$ we have $x^4+1 < \frac{-13}{2x+3}$. Thus according to Bolzano's Intermediate Value Theorem this equation have one root which is unique because of monotonically behaving functions.

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Exploiting Descartes law of signs is the way to go, anyway there is a (longer) alternative which consists in studying the variations of $f$ starting by its second derivative which is easily factorisable.

$f(x)=2x^5+3x^4+2x+16 $

$f'(x)=10x^4+12x^3+2=2(x+1)(5x^3+x^2-x+1)$

$f''(x)=40x^3+36x^2=4x^2(10x+9)$

So we can start draw a variation array

$\begin{array}{|c|ccccc|}\hline x & -\infty && -\frac 9{10} && 0 && +\infty\\\hline f'' & -\infty & \nearrow & 0 & \searrow & 0 & \nearrow &+\infty\\ && -&&+&&+\\\hline\end{array}$

$\begin{array}{|c|ccccc|}\hline x & -\infty && -1 && -\frac 9{10} && \alpha && +\infty\\\hline f' &+\infty &\searrow& 0 &\searrow & -0.187 & \nearrow & 0 &\nearrow& +\infty\\ &&+&&-&&-&&+&\\\hline\end{array}$

Since $f'(-\frac 9{10})<0$ and $\lim\limits_{x\to+\infty} f'(x)=+\infty$ by intermediate value theorem there is a root $f'(\alpha)=0$ in the interval $[-\frac 9{10},+\infty[$.

We don't need to calculate it, we just need to know that it annulates $g(x)=5x^3+x^2-x+1$.

$\begin{array}{|c|ccccc|}\hline x & -\infty && \beta && -1 && \alpha && +\infty\\\hline f &-\infty &\nearrow &0 &\nearrow & 15 &\searrow & f(\alpha) &\nearrow& +\infty\\\hline\end{array}$

Since $\lim\limits_{x\to-\infty}f(x)=-\infty$ and $f(-1)>0$ by intermediate value theorem there is a root $f(\beta)=0$ in the interval $]-\infty,-1]$.

To show it is the only one we have to prove that $f(\alpha)>0$.

The polynomial division of $f$ by $g$ gives $f(x)=\dfrac{50x^2+65x-3}{125}g(x)+\dfrac{2003+182x+18x^2}{125}$

Since $g(\alpha)=0$ then $f(\alpha)$ is the same sign as $2003+182\alpha+18\alpha^2$ but this quadratic has no real root so it is always positive and $f(\alpha)>0$.

You can eventually refine the interval for $\beta$ noticing $f(-2)=-4<0$ so $\beta\in]-2,-1[$.

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Just thought I'd give this as an answer, $$2(-x-1)^5 +3(-x-1)^4 +2(-x-1)+16=-2x^5-7x^4-8x^3-2x+15$$ there is only one sign change so by Descartes rule of signs, there is exactly one root of the original polynomial $x<-1$. On the other hand by fleablood's observation if $-1<x$ then $$0<-2+0-2+16<2x^5 +3x^4 +2x+16$$ Thus there is only one real root.

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