Proof involving Rolle's theorem and the MVT

Question

I'm stuck on a problem that asks me to show that the equation, $$6x^4-7x+1=0$$ does not have more than two distinct roots. I've tried to set up a proof by contradiction using Rolle's theorem and the MVT, but I can't arrive at a contradiction. I'm assuming that there are more than two distinct roots, specifically a third distinct root $c$, given distinct roots $a$ and $b$, and I'm trying to show that $c=b$ or $c=a$.

I would appreciate any advice/hints. Thanks.

Answer 1

If it has three distinct zeroes, say $a,b$ and $c$, then by Rolle's theorem the derivative should vanish in for at least one point in $(a,b)$ and at least one point in $(b,c)$. But the derivative is $$f^{'}(x)=24x^3-7.$$ and it can vanish at only one real number $\sqrt[3]{\dfrac{7}{24}}$.

Answer 2

I suppose you are talking about $x\in\mathbb{R}$. Assume that there are three distinct roots $a<b<c$. Then applying Rolle's theorem twice we know that there must be two distinct roots for the derivative: one root in $[a,b]$ and another in $[b,c]$. And how many distinct roots has the derivative?

Answer 3

If you know about the Intermediate value theorem, you can use the fact that you have $f(x)=6x^4-7x+1$, and that this function is continuous. Consider $f(-1)=6+7+1$, we have that $f(-1)>0$. now consider $f\Big(\frac{1}{2}\Big)=-\frac{17}{8}<0$. So the IVT say that for $f(a)\neq{f(b)}$, we have that $\forall\;y$ between $f(a)$ and $f(b)$, $\exists\;c\in\;]a,b[$ such as $f(c)=y$. We have that 0 is between $f(-1)$ and $f\Big(\frac{1}{2}\Big)$ so $\exists\;c\in]-1,\frac{1}{2}[$ such as $f(c)=0$. You can do the same for the fact that $0$ is between $f(2)$ and $f\Big(\frac{1}{2}\Big)$ so $\exists\;d\in\;]\frac{1}{2},2[$ such as $f(d)=0$.

We proved that there is two distinct roots. Now we must prove that there is only two roots. We can see if $f(x)$ is increasing for $x>2$.

\begin{equation} f(x+1)-f(x)=6(x+1)^4-7(x+1)+1-6x^4+7x-1\\ =6x^4+24x^3+36x^2+24x+6-7x-7-6x^4+7x\\ =24x^3+36x^2+24x-1>0,\;\forall\;x>0 \end{equation}

So $f(x)$ is always increasing.

So if we consider $M\in\mathbb{R}$ such as $M=f(x),\;\forall\;x\in]2,+\infty[$. We have that $0$ can't be between $f(2)$ and $f(M)$ and by the IVT, it's impossible to find $f(e)=0$ such as $e\in\;]2,M[$.

You can use the same strategy to prove that there is no other roots in the negative numbers.

This is your function.