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anonymous
 5 years ago
f(x) = x^3 + 2x + k. Prove that every function of this family has exactly one real root
anonymous
 5 years ago
f(x) = x^3 + 2x + k. Prove that every function of this family has exactly one real root

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of this function is olways positive and hence it has no real root, and so the function has only one real root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks, but you didn't prove that the function has roots or not ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the very fact that the derivative is positive and its an odddegree fn shows it has atleast one root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Answer: Let's assume that k represents any real number, let's say k = 3 f(x) = 3+2 x+x^3 Let's take two numbers : x = 30 and x = 12 Then, f(30) = 27057 < 0 f(12) = 1755 > 0 By the Intermediate Value Theorem, there exists a number "c" between 30 and 12 such that f(c) = 0. So, the equation has a real root. Now, let's suppose that there exists two roots w and q. Then, f(w) = f(q) = 0. However, using the Rolle's Theorem, f'(s) = 0 for some s in the set of numbers (w,q), but f'(x) = 2+3 x^2 > 0 for all x. So, it is impossible to have f'(s) = 0 for some c; in other words, there will be no maximum or minimum critical points, which means that there will be exactly one real root Therefore, we conclude that for any function of family f(x) = x^3 + 2x + k, where k is any real number, there exists exactly one real root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you can show that the derivative is positive for all values of x then the function is constantly increasing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have proven it that way. What do you think ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's basically what we were saying, but more long winded ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats extremely technical and purist but i think mine is a much easier and correct way...kudos 2 u for such theoristical capability

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did some research then I put my answer together.. Thanks all

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah uve said the same thing in a completely abstract and meticulous maner..great
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