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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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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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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