I was given the logistic growth equation $\frac{dN}{dt}=rN(1-\frac{N}{K})$ with $N(0)=N_o$.
I found the solution of this logistic equation: $N(t)=\frac{K}{\frac{K-N_o}{N_o}e^{-rt}+1}$.
Then, I was asked to show that the graph is concave up for $N_o < N < \frac{K}{2}$ and $N_o > K$ and that it is concave down for $\frac{K}{2} < N < K$.
I started by finding the first derivative of $N(t)$, however it is very messy and seems like there should be a more efficient way.
Any hint is appreciated!
$$\frac{dN}{dt}=rN(1-\frac{N}{K})\\N'=rN(1-\frac{1}{k}N)\\\text{ take derivative }\\N''=r(1-\frac{1}{k}N).N'+rN(0-\frac{1}{k}).N' \\\text{ plug N'}\\ N''=r(1-\frac{1}{k}N).rN(1-\frac{1}{k}N)+rN(0-\frac{1}{k}).rN(1-\frac{1}{k}N)=\\r^2N(1-\frac{1}{k}N)((1-\frac{1}{k}N)+N(0-\frac{1}{k}))= \\r^2N(1-\frac{1}{k}N)(1-\frac{2}{k}N)=\\\frac{r^2}{k^2}N(k-N)(K-2N)$$