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the derivative of this function is olways positive and hence it has no real root, and so the function has only one real root
thanks, but you didn't prove that the function has roots or not ?
the very fact that the derivative is positive and its an odd-degree fn shows it has atleast one root
Answer: 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
If you can show that the derivative is positive for all values of x then the function is constantly increasing.
I have proven it that way. What do you think ?
that's basically what we were saying, but more long winded ;)
thats extremely technical and purist but i think mine is a much easier and correct way...kudos 2 u for such theoristical capability
I did some research then I put my answer together.. Thanks all
yeah uve said the same thing in a completely abstract and meticulous maner..great