Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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




share|cite|improve this question
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
up vote 2 down vote accepted

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

share|cite|improve this answer









share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.