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?
Math question please Rolle theorem?
-
0Really interesting question... – 2012-12-26
4 Answers
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 $x$.
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.
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
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.