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I had this problem in an exam I recently appeared for:

Find the range of $$y =\frac{x^2+2x+4}{2x^2+4x+9}$$

By randomly assuming the value of $x$, I got the lower range of this expression as $3/7$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.

Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.

Another guy I met outside the exam center, told me he used an approach of $x$ tending to infinity in both cases and got the maximum value of this expression as $1/2$. But before I could ask him to explain more on this method, he had to leave for his work.

So, will someone please throw some light on this method of $x$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).

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    $\begingroup$ Use of calculus to find abslute maxima and minima is the easier way. $\endgroup$ Dec 1, 2021 at 6:39
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    $\begingroup$ I'll give a hint: An easy way to re-express the expression will be $\frac{1}{2}-\frac{1}{4\left(x+2\right)^{2}+14}$. Think of what circumstance would maximise the expression, which would be by minimizing the term subtracted. $\endgroup$
    – Prometheus
    Dec 1, 2021 at 6:41
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    $\begingroup$ @NikolaAlfredi Can you please show how. Because I am not versed with using the approach of calculus in such questions. $\endgroup$
    – Emanat S
    Dec 1, 2021 at 6:49
  • $\begingroup$ @EmanatS Try my approach, doesn't need calculus $\endgroup$
    – Prometheus
    Dec 1, 2021 at 6:50
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    $\begingroup$ The hint of Prometheus is very good. It gives the result quickly. $\endgroup$
    – kmitov
    Dec 1, 2021 at 6:51

4 Answers 4

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The question can be easily solved by this technique:

As $\displaystyle y = \frac {x^2 + 2x + 4}{2x^2 + 4x + 9} \implies 2y = \frac {2x^2 + 4x + 9 - 1}{2x^2 + 4x + 9}$.

Thus, $\displaystyle 2y = 1-\frac {1}{2(x + 1)^2 + 7} $

Squares can never be less than zero so the minimum value of the function : $\displaystyle 2(x + 1)^2 + 7 $ would be $7$ , or Maximum value of $\displaystyle \frac {1}{2(x + 1)^2 + 7} $ is $\displaystyle \frac {1}{7} $.

This tells that minimum value of $y $ will be $\displaystyle \frac{3}{7}$.

And so on.. check for $x \rightarrow \infty$.

From here you can easily tell the maximum and minimum values : $\displaystyle y \in \left [ \frac {3}{7}, \frac {1}{2} \right ) $

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  • $\begingroup$ Isn't very obvious on the first glance when it comes to the minimum, but to explain it, $2x^2+4x+9 = 2(x+1)^2+7$ $\endgroup$
    – Prometheus
    Dec 1, 2021 at 6:54
  • $\begingroup$ How did you split it just like that? Basically, how did you reduce that expression? $\endgroup$
    – Emanat S
    Dec 1, 2021 at 6:55
  • $\begingroup$ @Prometheus It's true... But I guess the test was a bit of a time-crusher, so perhaps the OP would like to get there in one step $\endgroup$
    – Spectre
    Dec 1, 2021 at 6:55
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    $\begingroup$ @Spectre Yes, I had to solve this problem under 50 seconds. And Prometheus, yes I know the method of completing the square. I just do not understand how to apply it to this problem. Please explain this approach to get a one step form like I'm five. $\endgroup$
    – Emanat S
    Dec 1, 2021 at 6:59
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    $\begingroup$ @EmanatS I have explained Nikola's step below, if you ever didn't get how he came to that simplification. $\endgroup$
    – Spectre
    Dec 1, 2021 at 7:04
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As a follow-up to @NikolaAlfredi's answer:

$ y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac{2x^2 + 4x + 8}{2(2x^2 + 4x + 9)} = \frac{2x^2+4x+9 - 1}{2(2x^2+4x+9)} = \frac{1}{2}(1-\frac{1}{2x^2+4x+9}) \implies 2y = 1 - \frac{1}{2x^2+4x+9}$. Now find the extremes of the range of the expression in the RHS of the above equation (which I believe you can; if not someone else or I myself shall try and add it) and divide them by $2$ to get the required extremes(taking half since we get values for $2y$ and not $y$).

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    $\begingroup$ Got it. I appreciate your kind efforts, and now it is easily clear to me. Thanks to both you, and Nikola. Also, can you please delve on calculus based approach like someone mentioned? I want to learn how to work this out with calculus too. $\endgroup$
    – Emanat S
    Dec 1, 2021 at 7:08
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    $\begingroup$ @EmanatS Are you looking for an answer using "calculus" methods? $\endgroup$ Dec 1, 2021 at 7:09
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    $\begingroup$ @TeresaLisbon Not really. But I am open to learning through it too. $\endgroup$
    – Emanat S
    Dec 1, 2021 at 7:09
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In general, if $\deg f = 0$ where $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$ the limit of $f$ as $x$ increases/decreases without bound is $a_n/b_n$.

In your case, $a_2 = 1$ and $b_2 = 2$. Hence, $a_2/b_2 = 1/2$.


We'll factor $f$ as $$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$

Notice that for all $x \in \mathbb{R}$, $f > 0$. Also, we can see that $(x+1)^2 + 2 < 2(x + 1)^2 + 7$. This means that the range should be a part of $(0,1/2)$. Since both numerator and denominator have $(x + 1)^2$ without any remaining $x$'s, we can see that this will be at its minimum when $x = -1$. Then, $$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$

Therefore, the range is $[3/7, 1/2)$.

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  • $\begingroup$ Note that both the numerator and denominator of $f$ must be polynomials with the same degree. This means that the expression could be linear/linear, quadratic/quadratic, cubic/cubic, and so on. $\endgroup$
    – soupless
    Dec 1, 2021 at 7:12
  • $\begingroup$ Woah, you're awesome. And so is @NikolaAlfredi I love this forum now, you both explained this problem so beautifully to me putting in all efforts to explain it to me. Thank you for a generalized solution to it as well. So, "limit of f as x increases/decreases without bound is an/bn" is always an upper limit of the expression right? Any places where I need to watch out this rule for? $\endgroup$
    – Emanat S
    Dec 1, 2021 at 7:17
  • $\begingroup$ Also for, (x^2+2x+1)/(4x^2-7x+9), a2/b2 will be {1/4}, but the upper limit of expression is 16/19. Shouldn't it be 1/4 according to your assertion? Edited $\endgroup$
    – Emanat S
    Dec 1, 2021 at 7:21
  • $\begingroup$ No. This is not always true. Consider the simple case $\frac{x}{x + 1}$. You'll see that it attains all real values except $y = 1$, and it is defined for all real $x$ except $x = -1$. In general, the limit is not the upper bound. It is only for the case where $p(x) < q(x)$ and $\deg p = \deg q$. ($p$ and $q$ are the polynomials in $f$, that is, $f(x) = p(x)/q(x)$). $\endgroup$
    – soupless
    Dec 1, 2021 at 7:23
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$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$

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