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 October 14th, 2018, 03:19 PM #1 Member   Joined: Sep 2011 Posts: 99 Thanks: 1 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. October 15th, 2018, 02:52 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 Every cubic equation has at most three real roots. October 15th, 2018, 07:14 PM   #3
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Quote:
 Originally Posted by skipjack 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)

Last edited by ProofOfALifetime; October 15th, 2018 at 07:22 PM. October 15th, 2018, 07:22 PM   #4
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 Originally Posted by Alexis87 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. Tags proof, rolle, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Nicol Calculus 2 March 11th, 2017 03:39 AM Azilips Calculus 4 March 17th, 2016 11:46 AM Shamieh Calculus 4 October 21st, 2013 04:18 PM mathkid Calculus 4 October 6th, 2012 06:35 PM mathkid Calculus 2 October 6th, 2012 07:17 AM

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