My Math Forum (http://mymathforum.com/math-forums.php)
-   Real Analysis (http://mymathforum.com/real-analysis/)
-   -   Proof using Rolle's Theorem (http://mymathforum.com/real-analysis/345136-proof-using-rolles-theorem.html)

 Alexis87 October 14th, 2018 03:19 PM

Proof using Rolle's Theorem

Hi, I have done up the proof for the question below. Please correct me if I have done wrong for the proof. Thanks in advanced!

Question: Prove that if ab < 0 then the equation ax^3 + bx + c = 0 has at most three real roots.

Proof:

Let f(x) = ax^3 + bx + c.

Assume that f(x) has 4 distinct roots, f(p) = f(q) = f(r) = f(s) = 0, there is a point x1 \in (p,q) such that f'(x1) = 0 ; x2 \in (q, r) such that f'(x2) = 0 ; x3 \in (r,s) such that f'(x3) = 0.

Since ab < 0 then there are two possibilities where a>0 and b<0 or a <0 , b > 0.

f'(x) = 3ax^2+b

If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 > 0 and b < 0, then f'(x) = 0

If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 < 0 and b > 0, then f'(x) = 0

This is not true because the equation f'(x) = 0 has only two roots.

Hence the given equation has at most three real roots when ab < 0.

 skipjack October 15th, 2018 02:52 AM

Every cubic equation has at most three real roots.

 ProofOfALifetime October 15th, 2018 07:14 PM

Quote:
 Originally Posted by skipjack (Post 600947) Every cubic equation has at most three real roots.
That's what I was thinking too when I saw the question. However, it needs to be proven (for practice)

 ProofOfALifetime October 15th, 2018 07:22 PM

Quote:
 Originally Posted by Alexis87 (Post 600934) Hi, I have done up the proof for the question below. Please correct me if I have done wrong for the proof. Thanks in advanced! Question: Prove that if ab < 0 then the equation ax^3 + bx + c = 0 has at most three real roots. Proof: Let f(x) = ax^3 + bx + c. Assume that f(x) has 4 distinct roots, f(p) = f(q) = f(r) = f(s) = 0, there is a point x1 \in (p,q) such that f'(x1) = 0 ; x2 \in (q, r) such that f'(x2) = 0 ; x3 \in (r,s) such that f'(x3) = 0. Since ab < 0 then there are two possibilities where a>0 and b<0 or a <0 , b > 0. f'(x) = 3ax^2+b If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 > 0 and b < 0, then f'(x) = 0 If the absolute value of 3ax^2 = the absolute value of b where 3ax^2 < 0 and b > 0, then f'(x) = 0 This is not true because the equation f'(x) = 0 has only two roots. Hence the given equation has at most three real roots when ab < 0.
You definitely have the right idea, (setting up contradiction) but there are some things that could be changed, one thing is at the beginning. You should define the roots first, then state f(p) = f(q) = f(r) = f(s) = 0.

Honestly, the steps that come after need some work. I'm super tired right now, but I'd be happy to look at this more tomorrow if someone else doesn't respond.

 All times are GMT -8. The time now is 06:19 AM.