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Calculus / general math related question

  1. May 24, 2015 #1
    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?
     
  2. jcsd
  3. May 24, 2015 #2

    pasmith

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    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].

    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].

    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].

    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.

    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.
     
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