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Prove $\;ax^3+bx+c=0\;$, with $\;a,b>0\;$, has at most one real root.

Should I use Rolle's theory for this? But I use it, what is the boundary of this function? I can only work out x must not be zero. Is the domain then turn into x is not equal to 0?

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  • $\begingroup$ plus, proving that has at most one real root also prove that there is exactly one real root :) $\endgroup$ – Ant Nov 14 '14 at 19:08
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The derivative of the function $f(x)=ax^3+bx+c$ is $$f'(x)=3ax^2+b$$ which is positive, so $f$ is strictly increasing and hence injective.

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If you really want to use Rolle's theorem, suppose to the contrary that there exist real roots $p$ and $q$ with $p\lt q$. Our function is certainly continuous in the interval $[p,q]$ and differentiable in $(p,q)$. So there exists a $t$ between $p$ and $q$ such that $f'(t)=0$. But $f'(t)=at^2+b\gt 0$.

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