Question

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)

Answer 1

This is an interesting question, have you gotten anywhere with it?

Answer 2

I'd say C-E are ruled out, since you can't make any assumptions about f(0) or g(0)

Answer 3

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)

Answer 4

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.

Answer 5

It's hard to think of two functions that don't adhere to A though

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

yeah, I guess so