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find the range for the expression, $f(n)=\frac{n^2+2\sqrt{n}(n+4)+4^2}{n+4\sqrt{n}+4}$ for $36 \le n \lt 72$

$f(n)=\frac{(\sqrt{n}+n+4)^{2}-9n}{(\sqrt{n}+2)^2}$,

$\sqrt{36}=6$

$\sqrt{72}=6\sqrt{2}$

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    @Simon Markett: yes it simplifies to,$f(n)=( n-2\sqrt{n} +4)$2012-08-20

2 Answers 2

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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 $\{n-2\sqrt n+4|36\leq n<72, n\in \mathbb N\}$.

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$f(n)=\frac{(\sqrt{n}+n+4)^2-9n}{(\sqrt{n}+2)^2}$

$f(n)=\frac{(\sqrt{n}+n+4)^2-(3\sqrt{n})^2}{(\sqrt{n}+2)^2}$

$f(n)=\frac{(\sqrt{n}+n+4-3\sqrt{n})(\sqrt{n}+n+4+3\sqrt{n})}{(\sqrt{n}+2)^2}$

$f(n)=\frac{(n+4-2\sqrt{n})(\sqrt{n}+n+4+3\sqrt{n})}{(\sqrt{n}+2)^2}$

$f(n)=\frac{(n+4-2\sqrt{n})(n+4+4\sqrt{n})}{(\sqrt{n}+2)^2}$

$f(n)=\frac{(n+4-2\sqrt{n})(\sqrt{n}+2)^2}{(\sqrt{n}+2)^2}$

$f(n)=(n+4-2\sqrt{n})$

$f(n)=(\sqrt{n}-2)^2+2\sqrt{n}$