PsychonautQQ said:
Let f and g be twice differentiable real-valued functions. If f'(x) > g'(x) for all values of x, which of the following statements must be true?
A) f(x) > g(x)
A counterexample is f(x) - g(x) = x. Then f'(x) - g'(x) = 1 > 0 for all x yet f(-1) - g(-1) = -1 < 0. Hence f(-1) < g(-1).
A counterexample is f'(x) = 1 and g'(x) = \tanh(x). Then f'(x) > g'(x) for all x, but f''(x) = 0 whilst g''(x) = \mathrm{sech}^2(x) > 0 for all x.
C) f(x) - f(0) > g(x) - g(0)
I don't think this one is actually true.
If f(0) = g(0) then this is exactly the same as (A), which was shown to be false. Indeed, the counterexample given for (A) also works here since if f(x) - g(x) = x then f(0) - g(0) = 0 so that f(0) = g(0).
What is true is that if f'(x) > g'(x) for all x then f(x) - f(0) > g(x) - g(0) for all
positive x.
D) f'(x) - f'(0) > g'(x) - g'(0)
Rearranging gives f'(x) - g'(x) > f'(0) - g'(0).
A counterexample is f'(x) - g'(x) = \mathrm{sech}(x): Then f'(x) - g'(x) > 0 for all x but f'(0) - g'(0) = 1 is maximal.
E) f''(x) - f''(0) > g''(x) - g''(0)
Rearranging, f''(x) - g''(x) > f''(0) - g''(0).
A counterexample is f'(x) - g'(x) = 1 + \tanh(x). Then f'(x) > g'(x) for all x. But f''(x) - g''(x) = \mathrm{sech}^2(x) so that f''(0) - g''(0) = 1 is maximal.
