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

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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Wolfram|alpha would sometimes do some good,you know – Zeta.Investigator Aug 20 '12 at 10:07
Try considering the function $g(x) = \frac{x^4 + 2 x (x^2+4)+4^2}{x^2+4x+4}$. It's easy to show it's strictly increasing, so $f(n) = g(\sqrt{n})$ is also strictly increasing. Hence, $f([36,72[) = [f(36),f(72)[$ by continuity. If $n$ is supposed to be a natural number, there's an easier solution: Just enumerate it. – Johannes Kloos Aug 20 '12 at 10:11
Try to simplify the expression first! Can you see the binomial formula? Does something cancel out? – Simon Markett Aug 20 '12 at 10:13
@Simon Markett: yes it simplifies to,$f(n)=( n-2\sqrt{n} +4)$ – Rajesh K Singh Aug 20 '12 at 10:20

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

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

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

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

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

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

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

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

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