# Math question please Rolle theorem?

I have to prove that the equation $$x^5 +3x- 6$$ can't have more than one real root..so the function is continuous, has a derivative (both in $R$) . In $R$ there must be an interval where $f'(c)=0$, and if I prove this,than the equation has at least one real root. So $5x^4+3 =0$ ..this equation is only true for $x=0$. How to prove that this is the only root?

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x=1.19334 when the equation is true –  mathguy Dec 26 '12 at 17:06
Note that you don't have to show that the equation $x^5+3x-6$ has a root; you only have to show it can't have more than one. So suppose it has more than one. This gives you the situation described in your second sentence (with the clause "and a $c$ in this interval" added). Now derive a contradiction. –  David Mitra Dec 26 '12 at 17:14
Really interesting question... –  user42912 Dec 26 '12 at 17:22

Note that the derivative is always positive. (It is not correct to say it is $0$ at $x=0$.)

Let $f(x)=x^5+3x-6$. If we had $a$ and $b$, with $a\ne b$, such that $f(a)=f(b)=0$, then by Rolle's Theorem there would be a $c$ between $a$ and $b$ such that $f'(c)=0$. But there cannot be such a $c$, since $f'(x)\gt 0$ for all $x$.

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Assume the function $f(x)=x^5+3x-6$ has two zeros or more. Let a and b be two zeros. Then Rolle's theorem tells you that there exists c between a and b such that $f'(c)=0$. Now $f'(c)=5c^4+3\geq 3$, a contradiction. So the function $f$ has at most one zero.

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So you want to prove $5x^4+3=0$ has only one root at $0$? That's not true as $5x^4+3>0$. This establishes the proof

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By Bolzano Theorem since $f(0)=-6$ and $f(2)=32$ we have $f(0)\cdot f(2)<0$ and then f has at least one root in $(0,2)$. Moreover, $f'(x)=5x^4+3>0$ so $f$ is strictly increasing and can only have one root.

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