# Concavity and Inflection Points

I have found a point called

$$\hat{t} = -\ln\left(-\ln\left(\frac{m}{P_0}\right) + 1\right)$$

$M$, and $P_0$ are constants, $M > P_0 > 0$

That when I plug it into my function $P(t)$ gives me an inflection point.

$$P(\hat{t}) = \frac{m}{e}$$

How do I find the concavity near the the inflection point? I am seeing pages online saying that I should basically find values that are before and after my $\hat{t}$, but it confuses me. What should I do to influence the value of $\hat{t}$ such that it shows whether $P(t)$ changes in concavity?

Note: $$P(t) = Me^{-ln(\frac{m}{P_0}) - e^{-kt}}$$

Edit: $k$ is also a constant

To understand which of the cases you are in, compute $P''(t)$. If you did everything correctly, $P''(m/e)$ should be zero. Look at what $P''(x)$ would be slightly to the left and to the right of $x=m/e$ and this will tell you the concavity -- $P''>0$ when in it concave up and $P''<0$ when it is concave down.
• I do get 0, did I compute $P(t)$ incorrectly? Doing this derivative I simplify $P(t) = -m(\frac{m}{P_0}) - e^{e^{-kt}}$ From this I basically multiply the entire equation by ln to get $P(t) = ln(-m(\frac{m}{P_0})) + e^{-kt}$. Then once more to get $P(t) = ln( ln(-m(\frac{m}{P_0}))) - kt$ The second derivative of this new equation is clearly 0, since $m$ and $P_0$ are constants. It does not look pretty, but is this legal – Hawaiian Rolls Apr 28 '17 at 12:13