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Math Help - Proof: f(x) > x^2 when f''(x) > 2 using MVT

  1. #1
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    Post Proof: f(x) > x^2 when f''(x) > 2 using MVT

    Some of the posters on this forum have been thoroughly impressive, so I'm hoping someone can help me out. I apologize for the presentation not being more professional, was getting "[LaTeX ERROR: Unknown error]" when trying to use Latex. Here's the problem as it was presented:

    Suppose f: R->R has first and second derivatives f' R->R and f'' R->R. Assume f(0) = 0, f'(0)=0, and f''(x)>2 for all x > 0. Show f(x) > x2 for all x > 0.
    Since if f(x) = x2 then f'(x) = 2x and f''(x) = 2 it seems clear that if f''(x) > 2, then f(x) > x2 . For the purposes of this problem we are to prove this using the mean value theorem. This was my approach:

    Consider the interval [0,x] for some x > 0. Using the MVT there exists some c such that f''(c) = f'(x) - f'(0) / x -0 = f'(x) / x. This is > 2 by assumption. Multiplying by x yields f'(x) > 2x.

    Next, again by MVT there exists some d such that f'(d) = f(x) - f(0) / x - 0 = f(x) / x. This is > 2x by the previous statement. Multiplying by x yields f(x) > 2x2 .

    However, I've been instructed this is incorrect. Apparently I'm misunderstanding the statements I'm making and that last line should be f(x) > 2cx but I'm not entirely sure why. Any assistance in explaining where I went wrong and/or how to approach this proof correctly would be greatly appreciated. Thanks in advance!
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  2. #2
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    Re: Proof: f(x) > x^2 when f''(x) > 2 using MVT

    Quote Originally Posted by ThomasNewton12 View Post
    Some of the posters on this forum have been thoroughly impressive, so I'm hoping someone can help me out. I apologize for the presentation not being more professional, was getting "[LaTeX ERROR: Unknown error]" when trying to use Latex. Here's the problem as it was presented:
    Since if f(x) = x2 then f'(x) = 2x and f''(x) = 2 it seems clear that if f''(x) > 2, then f(x) > x2 . For the purposes of this problem we are to prove this using the mean value theorem. This was my approach:
    Consider the interval [0,x] for some x > 0. Using the MVT there exists some c such that f''(c) = f'(x) - f'(0) / x -0 = f'(x) / x. This is > 2 by assumption. Multiplying by x yields f'(x) > 2x.
    Next, again by MVT there exists some d such that f'(d) = f(x) - f(0) / x - 0 = f(x) / x. This is > 2x by the previous statement. Multiplying by x yields f(x) > 2x2 .
    However, I've been instructed this is incorrect. Apparently I'm misunderstanding the statements I'm making and that last line should be f(x) > 2cx but I'm not entirely sure why. Any assistance in explaining where I went wrong and/or how to approach this proof correctly would be greatly appreciated. Thanks in advance!
    Just so you know, I have absolutely no idea what you have written means.

    But to use LaTeX here is an example.
    [tex] f(x)=x^2\text{ then }f'(x)2x [/tex] gives  f(x)=x^2\text{ then }f'(x)2x
    Click on the “go advanced” tab. On the toolbar you will see \boxed{\Sigma} clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.

    Your question seems to be about function concave up?
    But I cannot follow your notation.
    Thanks from topsquark and ThomasNewton12
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    Re: Proof: f(x) > x^2 when f''(x) > 2 using MVT

    Thanks for the quick response, Plato! Glad to see people willing to help. Sorry Latex wasn't working for me the first time, here's another attempt:

    Suppose f:R \rightarrow R has first and second derivatives f \prime :R \rightarrow R and f \prime \prime :R \rightarrow R. Assume f(0) = 0, f \prime (0) = 0, and f \prime \prime (x) > 2 for all x > 0.
    Show f(x) > x^2 for all x > 0.

    This seems like a simple proof. Note that if f(x) = x^2 then f \prime (x) = 2x and f \prime \prime (x) = 2, so it seems clear that if f \prime \prime (x) > 2 then  f (x) > x^2.

    Here was my approach:

    Consider the interval [0,x] for some x > 0. Using the mean value theorem, there exists a c \in (0,x) such that f \prime \prime (c) = \frac{f \prime (x) - f \prime (0)}{x-0} = \frac{f \prime (x)}{x}.

    Note since c > 0, \frac{f \prime (x)}{x} > 2 by assumption. Multiplying by x yields f \prime (x) > 2x for x > 0.

    Next again using the mean value theorem there exists a d \in (0, x) such that f \prime (d) = \frac{f(x) - f(0)}{x-0} = \frac{f(x)}{x}.

    Note since d > 0, \frac{f(x)}{x} > 2x by previous statement. Multiplying by x yields f(x) > 2x^2 for x > 0.

    However, as stated previously this is an invalid proof though I don't entirely understand why. I'd really appreciate understanding where I went wrong, and knowing how to approach this proof correctly. Hopefully this presentation is clearer!
    Last edited by ThomasNewton12; December 8th 2012 at 05:33 PM.
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