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anonymous
 3 years ago
Let f and g be twicedifferentiable realvalued 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)
anonymous
 3 years ago
Let f and g be twicedifferentiable realvalued 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is an interesting question, have you gotten anywhere with it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'd say CE are ruled out, since you can't make any assumptions about f(0) or g(0)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's hard to think of two functions that don't adhere to A though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0C 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 (fg)(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.
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