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.

  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 $f(x)=0.$

  • $\begingroup$ 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

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 $x$.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.