find the range for the expression, $f(n)=\frac{n^2+2\sqrt{n}(n+4)+4^2}{n+4\sqrt{n}+4}$ for $36 \le n \lt 72$
$f(n)=\frac{(\sqrt{n}+n+4)^{2}-9n}{(\sqrt{n}+2)^2}$,
$\sqrt{36}=6$
$\sqrt{72}=6\sqrt{2}$
find the range for the expression, $f(n)=\frac{n^2+2\sqrt{n}(n+4)+4^2}{n+4\sqrt{n}+4}$ for $36 \le n \lt 72$
$f(n)=\frac{(\sqrt{n}+n+4)^{2}-9n}{(\sqrt{n}+2)^2}$,
$\sqrt{36}=6$
$\sqrt{72}=6\sqrt{2}$
I decided to post this as an answer since you did the hardest bit yourself after my comment.
The function simplifies to $f(n)=n-2\sqrt n+4.$
We now want to find the intervals where $f$ is increasing or decreasing, respectively. We can do this either by differentiation:
$f'(n)=1-\frac{1}{\sqrt n}$
So $f$ is increasing in $[1,\infty)$.
Or by quadratic completion:
$f(n)=(\sqrt n-1)^2+3$
Again we conclude that $f$ is increasing in $[1,\infty)$. In particular the range of $f$ for $36\leq n<72$ will be $f(36)=28\leq f(n)<76-12\sqrt2$.
Addendum: If only natural numbers are allowed than you wont get anything more satisfactory than: the range is $\{n-2\sqrt n+4|36\leq n<72, n\in \mathbb N\}$.
$f(n)=\frac{(\sqrt{n}+n+4)^2-9n}{(\sqrt{n}+2)^2}$
$f(n)=\frac{(\sqrt{n}+n+4)^2-(3\sqrt{n})^2}{(\sqrt{n}+2)^2}$
$f(n)=\frac{(\sqrt{n}+n+4-3\sqrt{n})(\sqrt{n}+n+4+3\sqrt{n})}{(\sqrt{n}+2)^2}$
$f(n)=\frac{(n+4-2\sqrt{n})(\sqrt{n}+n+4+3\sqrt{n})}{(\sqrt{n}+2)^2}$
$f(n)=\frac{(n+4-2\sqrt{n})(n+4+4\sqrt{n})}{(\sqrt{n}+2)^2}$
$f(n)=\frac{(n+4-2\sqrt{n})(\sqrt{n}+2)^2}{(\sqrt{n}+2)^2}$
$f(n)=(n+4-2\sqrt{n})$
$f(n)=(\sqrt{n}-2)^2+2\sqrt{n}$