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Let f and g be twice-differentiable real-valued functioned defined on R. If f'(x) > g'(x) for all x > 0, which of the following inequalities must be true for all x > 0 ? (A) f(x) > g(x) (B) f''(x) > g''(x) (C) f(x) - f(0) > g(x) - g(0) (D) f'(x) - f'(0) > g'(x) - g'(0) (E) f''(x) - f''(0) > g''(x) - g''(0)

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This is an interesting question, have you gotten anywhere with it?
I'd say C-E are ruled out, since you can't make any assumptions about f(0) or g(0)
Actually, nevermind, you can. Because if f'(x) > g'(x) very close to zero, the definition of differentiability seems like it would guarantee that f'(0) > g'(0)

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I have not looked at the problem yet myself. I posted this as a challenge. The question is saying that, after x = 0, f increases faster than g. We certainly can eliminate A because this says nothing of the values of f(x) or g(x). I am learning towards C only because of what I imagine the function does at the point x = 0. Taking into account your previous post, it appears we are in agreement.
It's hard to think of two functions that don't adhere to A though
f(x) = x^3 - 10 g(x) = x^2 Here, f(1) < g(1) so A fails. But f'(x) > g'(x) for all x > 0
yeah, I guess you're right, so are we going with C or D? Because if it's twice differentiable, then D sounds like a better answer to me
C says f(x) - f(0) > g(x) - g(0) D says f'(x) - f'(0) > g'(x) - g'(0) given: f'(x) > g'(x), f'(x) - g'(x) > 0 We can write this as a composite function (f - g)'(x) > 0 This means that the composite function (f-g)(x) is increasing. Which equivalently means f(x) - g(x) > 0 This is why I think the answer is C. The logic isn't sound but the line of reasoning seems ok.
yeah, I guess so

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