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:

Quote:

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) > x^{2} for all x > 0.

Since if f(x) = x^{2} then f'(x) = 2x and f''(x) = 2 it seems clear that if f''(x) > 2, then f(x) > x^{2} . 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) > 2x^{2} .

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!

Re: Proof: f(x) > x^2 when f''(x) > 2 using MVT

Quote:

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) = x^{2} then f'(x) = 2x and f''(x) = 2 it seems clear that if f''(x) > 2, then f(x) > x^{2} . 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) > 2x^{2} .

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 $\displaystyle f(x)=x^2\text{ then }f'(x)2x $

Click on the “go advanced” tab. On the toolbar you will see $\displaystyle \boxed{\Sigma}$ 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.

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

Show $\displaystyle f(x) > x^2$ for all $\displaystyle x > 0$.

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

Here was my approach:

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

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

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

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

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!