Calculus / general math related question

[PsychonautQQ]

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)
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)

The correct answer is C. Can somebody explain to me why the other answers don't work and why C works?

 

[pasmith]

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 [itex]f(x) - g(x) = x[/itex]. Then [itex]f'(x) - g'(x) = 1 > 0[/itex] for all [itex]x[/itex] yet [itex]f(-1) - g(-1) = -1 < 0[/itex]. Hence [itex]f(-1) < g(-1)[/itex].

B) f''(x) > g''(x)

A counterexample is [itex]f'(x) = 1[/itex] and [itex]g'(x) = \tanh(x)[/itex]. Then [itex]f'(x) > g'(x)[/itex] for all [itex]x[/itex], but [itex]f''(x) = 0[/itex] whilst [itex]g''(x) = \mathrm{sech}^2(x) > 0[/itex] for all [itex]x[/itex].

C) f(x) - f(0) > g(x) - g(0)

I don't think this one is actually true.

If [itex]f(0) = g(0)[/itex] 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 [itex]f(x) - g(x) = x[/itex] then [itex]f(0) - g(0) = 0[/itex] so that [itex]f(0) = g(0)[/itex].

What is true is that if [itex]f'(x) > g'(x)[/itex] for all [itex]x[/itex] then [itex]f(x) - f(0) > g(x) - g(0)[/itex] for all positive [itex]x[/itex].

D) f'(x) - f'(0) > g'(x) - g'(0)

Rearranging gives [itex]f'(x) - g'(x) > f'(0) - g'(0)[/itex].

A counterexample is [itex]f'(x) - g'(x) = \mathrm{sech}(x)[/itex]: Then [itex]f'(x) - g'(x) > 0[/itex] for all [itex]x[/itex] but [itex]f'(0) - g'(0) = 1[/itex] is maximal.

E) f''(x) - f''(0) > g''(x) - g''(0)

Rearranging, [itex]f''(x) - g''(x) > f''(0) - g''(0)[/itex].

A counterexample is [itex]f'(x) - g'(x) = 1 + \tanh(x)[/itex]. Then [itex]f'(x) > g'(x)[/itex] for all [itex]x[/itex]. But [itex]f''(x) - g''(x) = \mathrm{sech}^2(x)[/itex] so that [itex]f''(0) - g''(0) = 1[/itex] is maximal.