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Show equation has Exactly one real root 
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#1
Jul608, 09:37 PM

P: 66

1. The problem statement, all variables and given/known data
show that the equation 1 + 2x + x^3 + 4x^5 = 0 has exactly one real root 2. Relevant equations 3. The attempt at a solution i dont know what to use to find out that it has exactly one real root. i know you use the intermediate value theorem for roots but what do you do to show it has exactly one? 


#2
Jul608, 09:55 PM

P: 152

First show that it has at least one real root (you can use the intermediate value theorem, as you noted). Then consider its derivative. What does it tell you about this function?



#3
Jul608, 10:26 PM

P: 66

i used f(1) and f(1) to prove the intermediate value theorem. the derivative of the function is 2+3x^2+20x^4 which i guess shows that the function is always positive?



#4
Jul608, 11:38 PM

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Show equation has Exactly one real root



#5
Nov2008, 07:30 PM

P: 1

Take two numbers, say 1 and 0. Then f(1) = 6 < 0 and f(0) = 1 > 0 (Letting f(x) = 1 + 2x + x^3 + 4x^5.)
By the Intermediate Value Theorem there exists a number "c" between 1 and 0 such that f(c) = 0. So the equation has a real root. Then suppose there are two roots a and b. Then f(a) = f(b) = 0 and by Rolle's Theorem f'(r) = 0 for some r in the set of numbers (a,b). But f'(x) = 2 + 3x^2 + 20x^4 > 0 for all x, so it is impossible to have f'(r) = 0 for some c. So there is exactly one root. 


#6
Apr309, 12:52 AM

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#7
Apr309, 05:27 AM

P: 336




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