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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 ).

amWhy
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Emanat S
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    Use of calculus to find abslute maxima and minima is the easier way. – Nikola Alfredi Dec 01 '21 at 06:39
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    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. – Prometheus Dec 01 '21 at 06:41
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    @NikolaAlfredi Can you please show how. Because I am not versed with using the approach of calculus in such questions. – Emanat S Dec 01 '21 at 06:49
  • @EmanatS Try my approach, doesn't need calculus – Prometheus Dec 01 '21 at 06:50
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    The hint of Prometheus is very good. It gives the result quickly. – kmitov Dec 01 '21 at 06:51
  • @Prometheus I honestly loved your approach, but the issue is I can not reduce it down to this form quickly at the first glance. You need to work your way out for a few seconds, and in a test taking format that becomes hard. Will you please shed some light on how to use calculus based approach, and then I can figure out what works out the best for me. Thanks. – Emanat S Dec 01 '21 at 06:52
  • @EmanatS Check out my solution. It doesn't use calculus but the property of Rational function and quadratics. – Nikola Alfredi Dec 01 '21 at 06:53
  • It seems that you have not learned calculus, which is why I don't want to recommend it for your usage. Plus the derivation of the expression is even more torturous. Also I meant $x+1$ in my comment, not $x+2$. – Prometheus Dec 01 '21 at 06:53
  • @Prometheus Isn't there a standard set of calculus based generalization for such quadratic / quadratic forms? I do know differentiation though, kinda rusty on integration. – Emanat S Dec 01 '21 at 06:57
  • Another way, which works for fractions of quadratics in general and does not use calculus. Rewrite the equation as a quadratic in $,x,$ by eliminating the denominator and regrouping: $y =\frac{x^2+2x+4}{2x^2+4x+9}$ $\iff (2x^2+4x+9)y=x^2+2x+4$ $\iff (2y-1)x^2+2(2y-1)x+9y-4=0,$. The range of $,y,$ is the range for which the quadratic has real roots i.e. $\frac{1}{4} \Delta$ $= (2y-1)^2 - (2y-1)(9y-4)$ $=(2y-1)(-7y+3) \ge 0,$. – dxiv Dec 01 '21 at 07:26
  • @dxiv I know this method, but it's too lengthy for a test taking format where you've to solve this problem under 50 seconds. So, I was looking for a faster method. – Emanat S Dec 01 '21 at 07:27
  • @EmanatS Compare it to the other methods for a slightly modified $,y =\frac{x^2+\color{red}{3}x+4}{2x^2+4x+9},$. – dxiv Dec 01 '21 at 07:35

5 Answers5

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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 ) $

Martin Sleziak
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Nikola Alfredi
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  • Isn't very obvious on the first glance when it comes to the minimum, but to explain it, $2x^2+4x+9 = 2(x+1)^2+7$ – Prometheus Dec 01 '21 at 06:54
  • How did you split it just like that? Basically, how did you reduce that expression? – Emanat S Dec 01 '21 at 06:55
  • @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 – Spectre Dec 01 '21 at 06:55
  • @EmanatS Have you learnt completing the square? – Prometheus Dec 01 '21 at 06:56
  • @Prometheus , Yes it is indeed very obvious but I guess he is a new learner. So rather I explained him this way. – Nikola Alfredi Dec 01 '21 at 06:57
  • @EmanatS Multiply $2$ on both sides and then you can already see it. – Nikola Alfredi Dec 01 '21 at 06:58
  • I kinda meant that the reasoning doesn't look exactly clear, which might mean that they won't understand why we get the $\frac{3}{7}$ – Prometheus Dec 01 '21 at 06:58
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    @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. – Emanat S Dec 01 '21 at 06:59
  • @Prometheus Oh ok... he mentions the 'Quadratics' in his tags So I though he would be aware about the Vertex co-ordinates, that here give the minima of quadratic equation. – Nikola Alfredi Dec 01 '21 at 07:00
  • @EmanatS I will add a few lines to explain it. – Nikola Alfredi Dec 01 '21 at 07:01
  • I know that minima and maxima of a quadratic lies at x= -b / 2a. This problem could've been easily figured out by multiplying it with 2 and then working out the maths from there on. But what if there wasn't any way to make the numerator and denominator kinda same like that? Does there exists any standard and quick way to get a range of any quadratic / quadratic expression? – Emanat S Dec 01 '21 at 07:03
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    @EmanatS I have explained Nikola's step below, if you ever didn't get how he came to that simplification. – Spectre Dec 01 '21 at 07:04
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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 $y$).

Spectre
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    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. – Emanat S Dec 01 '21 at 07:08
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    @EmanatS Are you looking for an answer using "calculus" methods? – Sarvesh Ravichandran Iyer Dec 01 '21 at 07:09
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    @TeresaLisbon Not really. But I am open to learning through it too. – Emanat S Dec 01 '21 at 07:09
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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 $[3/7, 1/2)$.

soupless
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  • 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. – soupless Dec 01 '21 at 07:12
  • 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? – Emanat S Dec 01 '21 at 07:17
  • 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 – Emanat S Dec 01 '21 at 07:21
  • 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, $f(x) = p(x)/q(x)$). – soupless Dec 01 '21 at 07:23
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$$y = \frac{x^2 + 2x + 4}{2x^2 + 4x + 9} = \frac12 - \frac {1/2}{2x^2 + 4x + 9} \implies \frac {dy}{dx} = \frac {4x + 4}{\text{whatever}} \text{ Let } \color{green}{\frac{dy}{dx} = 0 \implies x = -1}, y(x = -1) = \color{blue}{\frac37} \text{ also } y(x\to \infty) = \color{blue}{\frac 12}$$

Darshan P.
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Special Case for This Problem

You can simplify through substitution $u=(x+1)^{2}$:

$$ \begin{aligned} y = \frac{x^{2}+2x+4}{2x^{2}+4x+9} &= \frac{u+3}{2u+7} \\\\ &\in \left[\frac{3}{7},\frac{1}{2}\right) \phantom{x} \forall \phantom{x} u\geq 0 \end{aligned} $$

You can see on the right, as $u$ grow larger and larger the expression approaches $\frac{1}{2}$. This is the method that your friend mentioned.

General Case

After making sure that the denominator is non-zero, do algebra manipulation like the following.

$$ y = \frac{x^{2}+2x+4}{2x^{2}+4x+9} \implies (2y-1)x^{2}+(4y-2)x+9y-4 = 0 $$

In order to have solution, the discriminant must be non-negative.

$$ \begin{aligned} 0 &\leq (4y-2)^{2} - 4(2y-1)(9y-4) \\ &= -56y^{2}+52y-12 \\ &= -(7y-3)(8y-4), \phantom{x} y\neq \frac{1}{2} \end{aligned} $$

One can see that $y\in\left[\frac{3}{7},\frac{1}{2}\right)$.

acat3
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