# Proving a function is below zero using the mean value theorm

Let $\varphi (w)\Psi (x)$ be a solution to the heat problems. Given that $\Psi (-1)=\Psi (1)=0$ prove that it is not possible for $\Psi (x)$ and $\Psi ''(x)$ to be strictly positive for all $x$ such that $−1 < x < 1$. You may use any facts, such as the mean value theorem.

So yes, this is homework. But we have been at it for hours now,and could use a little nudge.

Assume $\Psi (x),\Psi ''(x)$ are positive. Then the mean value theorem states $\Psi '(c) = 0$ according to:

\begin{equation*} \Psi'(c) = \frac{\Psi (1)-\Psi (−1)}{1+(-1)}. \end{equation*}

Since $\Psi (c) = 0$, then we know there exists a tangent parallel to $x-$axis. This is the extrema which is strictly positive. Here is where we get stuck.

Since we assumed $\Psi ''(c)$ is positive then we know that there exists a minimum due to the the derivative test. But how does this help us prove that the range from $-1$ to $1$ is NOT strictly positive.

A little background info: This is for real analysis. Both me and roommate have not taken a prior proofs course (which is actually the pre-req). So we are not experts are proving so if any of you take the time to answer, please dumb it down as much as you can.

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Hint:

If $\psi(x)$ is strictly positive, what does it say about the trend of $\psi'(x)$?

If $\psi''(x)$ is strictly positive, what does it say about the trend of $\psi'(x)$?

It might be useful to draw a graph.

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sorry, what does it mean by the trend. we are trying to draw graphs, but don't know because we dont have an equation to draw. – Tyler Hilton Sep 23 '10 at 21:41
@Affan: Just draw randomly (use your imagination), keeping $\psi(-1) = \psi(1) = 0$ and $\psi(x)$ strictly positive. – Aryabhata Sep 23 '10 at 21:48
trend here means tendency, or what do you know about the derivative's behavior based on the behavior of its antiderivative and its derivative. He means draw some graphs for functions you know and look at how the relationships he points out behave geometrically. @Moron, if I have misrepresented your answer I am sorry, I was just trying to help. – BBischof Sep 23 '10 at 21:49
@BBischof: No need to apologise :-) – Aryabhata Sep 23 '10 at 21:51
sorry, very stupid question but what does it mean when someone says $ψ(x)$ is strictly positive ? does it mean the entire graph (or interval we are working with) is above the x axis? – Tyler Hilton Sep 23 '10 at 21:52