Thread: 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!

Originally Posted by ThomasNewton12

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
Click on the “go advanced” tab. On the toolbar you will see 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 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 has first and second derivatives and . Assume , , and for all .
Show for all .

This seems like a simple proof. Note that if then and , so it seems clear that if then .

Here was my approach:

Consider the interval for some . Using the mean value theorem, there exists a such that .

Note since , by assumption. Multiplying by x yields for .

Next again using the mean value theorem there exists a such that .

Note since , by previous statement. Multiplying by x yields for .

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!