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I have deduced a proof as stated below and am not sure if it is correct, therefore need some advice.

Proof: Let $f(x) = ax^3+bx+c = 0$. $f(x)$ is continuous since it is a polynomial and it is differentiable since it has a limit. Assume $f(x)$ has $2$ roots, $f(a) = 0$ and $f(b) = 0$, there is a point $d\in(a,b)$ such that $f'(d) = 0$.

$$f'(x) = 3ax^2+b $$ Since $ab>0$, $a$ and $b$ must be both positive or both negative. $f'(d) = 3a(d)^2+b = 3ad^2+b$ not equal to $0$ instead $>0$ since any values of $d$ for $d^2$ will be positive.

Likewise $f'(d) = 3a(d)^2 + b$ will not equal to $0$ instead $<0$ for all negative values of $a$ and $b$.

Hence, a contradiction, $f$ has exactly one root.

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  1. You should begin the proof with "Let $f(x)=ax^3+bx+c$". It is commonly understood that any reader should understand that this means "Let $f(x)=ax^3+bx+c$ for all $x$ (i.e.for all $x\in \Bbb R$ )"...rather than writing "$f(x)=ax^3+bx+c=0$".

  2. You should not use $a,b$ for possible values of $x$ for which $f(x)=0 $ because the letters $a,b$ are already being used for something else. "Assume $f(x_1)=f(x_2)=0$ with $x_1<x_2.$ Then by Rolle's Theorem there exists $d\in (x_1,x_2)$ such that $f'(d)=0$."

  3. Allowing for the poor grammar in the rest of the proof, it is correct EXCEPT for the last line. You have shown that there is at MOST one $x$ such that $f(x)=0.$ But you have not shown that there is also at LEAST one $x$ such that $f(x)=0.$

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  • $\begingroup$ Thanks noted on that. $\endgroup$ – Alexander Oct 15 '18 at 7:45
  • $\begingroup$ This is a good answer addressing OP's question instead of solving OP's problem. $\endgroup$ – Surb Feb 21 '19 at 0:14
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The inequality after first derivative can be proven with discriminant:

$$\begin{align*} f'(x) &= 3ax^2 + b\\ D &= 0-4\cdot(3a)\cdot b\\ &< 0 \end{align*}$$

So $f'(x) \ne 0$ for all $x$.

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    $\begingroup$ Does the OP prove that $f$ has any real roots? $\endgroup$ – Michael Burr Oct 14 '18 at 10:22
  • $\begingroup$ @MichaelBurr. No. He did not. I pointed that out in my answer. $\endgroup$ – DanielWainfleet Oct 14 '18 at 10:33
  • $\begingroup$ Thanks noted on that. $\endgroup$ – Alexander Oct 15 '18 at 7:45

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