MHB Using Rolle's theorem to prove for roots

[Joe20]

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

Question:
Prove that if ab > 0 then the equation ax^3 + bx + c = 0 has exactly one root by rolle's theoremProof:
Let f(x) = ax^3+bx+c = 0. f(x) is continuous and differentiable since it is a polynomial.

Assume f(x) has 2 roots, f(a) = 0 and f(b) = 0, there is a point d element of (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.

 

[Olinguito]

Re: Using rolle's theorem to prove for roots

Your proof is essentially correct (although in one line you used $a$ and $b$ as roots of the polynomial whereas they’re already used as coefficients of the polynomial).

Since $ab>0$, $a$ and $b$ are either both positive or both negative; therefore $f'(x)=3ax^2+b$ is either always positive or always negative, i.e. it is never zero. If $f(x)$ had two distinct real roots, Rolle’s theorem would imply there was a point between the roots at which the derivative was zero – a contradiction. Hence the polynomial cannot have more than one real root.

Finally, note that the polynomial is cubic; therefore it must have at least one real root. Conclusion: it has exactly one real root.

 

[Joe20]

Re: Using rolle's theorem to prove for roots

Your proof is essentially correct (although in one line you used $a$ and $b$ as roots of the polynomial whereas they’re already used as coefficients of the polynomial).

Since $ab>0$, $a$ and $b$ are either both positive or both negative; therefore $f'(x)=3ax^2+b$ is either always positive or always negative, i.e. it is never zero. If $f(x)$ had two distinct real roots, Rolle’s theorem would imply there was a point between the roots at which the derivative was zero – a contradiction. Hence the polynomial cannot have more than one real root.

Finally, note that the polynomial is cubic; therefore it must have at least one real root. Conclusion: it has exactly one real root.

Thanks, will change the usage of a and b accordingly.