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I'm so, so very confused about finding the ranges of real functions, no concept in Mathematics has yet confused me more than this, please tell me what's wrong in my solution for finding the range of the function : $\dfrac{3}{2-x^2}$

Here's how I do it and get a partial answer, please check it out...

$x^2 \geq 0$
$-x^2 \leq 0$
$2 - x^2 \leq 2$
$\dfrac {1}{2 - x^2} \geq \dfrac{1}{2}$
So, $\dfrac {3}{2 - x^2} \geq \dfrac{3}{2}$
So, $f(x) \geq \dfrac{3}{2}$

By this, $Range(f) = [\dfrac{3}{2}, ∞)$

But as per my textbook, the answer is $(-∞,0)∪[\dfrac {3}{2},∞)$, which is (obviously) correct

What my main question here is : How can I add the proof of the negative values in the range in my proof?

I would be very, very grateful to you if you help (no exaggeration, I would be so very thankful cause this topic is frustrating me)

Also, this is a general question : Am I the only one so confused about finding domains and ranges? I mean did you, when you began, also face problems with this concept?

Thanks

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6 Answers 6

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The problem comes when you take the reciprocal. The step $2-x^2 \leq 2$ is correct. When you invert, however, you have to look at 2 cases; when $2-x^2 \geq 0$ and $2-x^2 \leq 0$. When you take the reciprocal, you assume that the first case is correct. The second case, however, gives rise to the other part of the range.

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  • $\begingroup$ Thanks, but I'd be thankful if you can tell me the full new solution and a method to find out the range for any given function, I am so, so, so confused about this, every single time I get the wrong answer $\endgroup$ Apr 7, 2020 at 12:33
  • $\begingroup$ @Rajdeep_Sindhu, you were on the right track. It's just with inequalities you have to be careful to account for all the possibilities. $\endgroup$
    – Paul
    Apr 7, 2020 at 12:37
  • $\begingroup$ Also, can you solve relation for every function like : find a relationship between x and some constant from the function, like if the function is : $f(x) = \sqrt {x-}$, then the condition that we have is $x - 5 \geq 0$, so we can then evaluate this inequality to obtain $f(x)$ and then find the range taking care of other conditions or shall I use this method : equate $f(x) = y$ and then find the inverse function and it's domain which will then be equal to the range of the original function. I apologize if I'm being irritating but I'm really fed up now. $\endgroup$ Apr 7, 2020 at 12:40
  • $\begingroup$ Okay, got it, thanks again! $\endgroup$ Apr 7, 2020 at 12:43
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There was an error in operating with inequalities:

$2 - x^2 \le 2$ implies $\dfrac {1}{2 - x^2} \geq \dfrac{1}{2}\;$ only if both sides have the same sign – positive here.

Other than that, if $2 - x^2 < 0$, its inverse remains negative, so it is less than $\frac12$.

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When you write : $$ 2 - x^2 \leq 2 \implies \frac 1{2 - x^2} \geq \frac 12 $$

you ignore the fact that $2-x^2$ could be negative. Therefore, you must do as follows : $$ 2 - x^2 \leq 2 \implies \begin{cases} \frac{1}{2 - x^2} \in (-\infty, 0] & 2 - x^2 < 0 \\ \frac 1{2-x^2} \in [\frac 12 , \infty) & 2 \geq 2-x^2 > 0 \end{cases} $$

Having done this, from the analysis of the lower case you get your range.

From the analysis of the upper case, you get by multiplication by a positive number, the same range. Hence the answer.


Always , while computing ranges, ensure that you see the sign of what is being multiplied or divided by in an expression, very carefully. The fact that $2-x^2$ could be negative was ignored, leading to the issue.

I am confident that if you attempt about ten of these kind of problems, you will feel more confident. This was the case for me. Usually textbooks will drill these kind of issues into you, and you will see fifteen examples strewn all over. Do all of them, and you won't make a mistake again.

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  • $\begingroup$ Additionally, see similar questions on the site for model solves of the kind of questions you are finding difficult i.e. domain / range questions. $\endgroup$ Apr 7, 2020 at 12:34
  • $\begingroup$ Thanks, so basically I just have to take care of all the factors, exceptions etc. and I will make it, right? $\endgroup$ Apr 7, 2020 at 12:42
  • $\begingroup$ You will make it with flying colours. Which textbook are you taking questions out of? Additionally, just searching questions of domain/range with the tag elementary set theory will give you good examples on this site. $\endgroup$ Apr 7, 2020 at 12:44
  • $\begingroup$ RD Sharma Class 11 Mathematics. It's an Indian textbook, more like 'the' Indian textbook, since most of the students use it in India... $\endgroup$ Apr 7, 2020 at 12:46
  • $\begingroup$ It is "the" textbook? Then it must have plenty of examples. Do them all, post those you struggle with (keeping format of site in mind : this question was fantastic) and you will be through this one in ten days maximum. $\endgroup$ Apr 7, 2020 at 12:48
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Let's define $f(x) = \frac{3}{2 - x^2}$. It is true that $2 - x^2 \leq 2$, but you need to discuss the case where $2 - x^2 = 0$, i.e. $x = \pm \sqrt{2}$, that's the part where you forgot to discuss. In particular, as $x \rightarrow \sqrt{2}^+$, $f(x) \rightarrow -\infty$ and as $x \rightarrow \pm \infty$, $f(x) \rightarrow 0$ from below hence $0$ is not inclusive in the range. Finally, the function is even so that makes you job slightly easier.

Generally speaking, you need to determine the domain of a function properly to find its range.

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Another way to find the range is to solve for $x$. Thus put $y = \dfrac{3}{2-x^2}\implies 2y-x^2y=3\implies x^2 = \dfrac{2y-3}{y}$. So in order for $x$ to exist for each value of $y$ in the range, we must find the conditions on $y$ and the range of the function will come out from these conditions. First $x^2 \ge 0$ and $x^2 \neq 2$ mean that $\dfrac{2y-3}{y} \ge 0$ and $\dfrac{2y-3}{y} \neq 2$. The second inequality already holds for any $y$ while the first inequality holds if $y < 0$ or $y \ge \frac{3}{2}$ and this is precisely the range of the function. In summary: $\text{Range}(f) = (-\infty, 0)\cup [\frac{3}{2}, \infty)$.

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  • $\begingroup$ That's a good method to do it but how did you use $\dfrac {2y-3}{y} \geq 0$ to figure out that $y < 0$ or $y \geq \dfrac {3}{2}$? An answer to this might just clear all of my doubts. $\endgroup$ Apr 7, 2020 at 13:34
  • $\begingroup$ Because $\dfrac{2y-3}{y} = x^2 \ge 0$ $\endgroup$
    – DeepSea
    Apr 7, 2020 at 13:36
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We have: $$y=f(x)=\frac{3}{2-x^2} ; x \ne \pm\sqrt{2}$$ Write $x$ as a function of $y$ as: $$x=\sqrt{2-\frac{3}{y}}$$ Now we have to find the set of values of $y$ for which $x$ is $\mathbf {real}$.

For this, $2-{3\over y} \ge 0$. This can be acquired in two ways:

$\mathbf {Case\ 1:}$ Substituting $x=0$ into the second equation we get $y={3\over 2}$. Hence talking of positive $y$, it can go from $3\over 2$ all the way upto $\infty$, this is because no matter how large the value of $y$ you take, $x$ will always be a little less than $\sqrt 2$, which is safe. $y$ cannot be less than $3\over 2$ because that would mean $x$ is imaginary. $\mathbf {So},\ \mathbf {y\in [{3\over 2}, \infty)}$

$\mathbf {Case\ 2}:$ We note that from the above step, we get $x\in [0,\sqrt 2)$, when $y\in [{3\over 2},\infty)$. But a little observation shows us that if $y\lt 0$, the whole expression $(2-{3\over y})$ can and will take every value $\mathbf {between}$ $2$ and $\infty$, as $y$ goes from $-\infty$ to $0$ (but never equal to $0$), thus giving $x\in (\sqrt 2,\infty)$.

Thus we have $range(f)= (-\infty,0)\cup[{3\over 2}, \infty)$ that gives us real $x$.

There is no need to care about the other possble values of $x$ as our main focus is $range(f)$ i.e. we need values of $y$ as expressed in the second equation.

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