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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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    @Affan: I suggest you edit the question and add an update at the end to show exactly what you are thinking. From your comment it is not clear what you mean.2010-09-23