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Have a cubic with no stationary points which is obviously increasing for all values of x, can't think how to prove it ><
I'm sure the proof is very obvious
I can do basic differentiation etc.
Thanks
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You have a function f(x). What you must prove is that if a < b, then f(a) < f(b). But if you're doing calculus, an alternate proof would be that f'(x) > 0 for all values of x.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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YBut if you're doing calculus, an alternate proof would be that f'(x) > 0 for all values of x.
I understand that the gradient function must always be positive... but how do you prove that?
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You know its a cubic, so the gradient function must be in the form of a quadratic. You also know that this quadratic never passes through y = 0. What can you say about such a quadratic function?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Have a cubic with no stationary points which is obviously increasing for all values of x
This is not true. The graph y = −x[sup]3[/sup]−x has no stationary points but its decreasing.
Last edited by JaneFairfax (2007-06-10 21:34:50)
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The way I read it was "I have a cubic with not stationary points, and it's obvious by looking at it that it's increasing."
If that's the case, then that first term has to be positive.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Thats right. A cubic with no stationary points in which the coefficient x[sup]3[/sup] is positive is increasing. But a cubic simply with no stationary points doesnt have to be increasing its decreasing if the coefficient x[sup]3[/sup] is negative.
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