# 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

## 1 Answer

HINT Inflection point is a place where concavity changes. That means before the inflection point, the concavity is positive and after it, is negative, or vice versa.

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.

• That's one long derivative. Here I go :) – Hawaiian Rolls Apr 28 '17 at 12:05
• 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
• @HawaiianRolls yes it is correct – gt6989b Apr 28 '17 at 17:48