1
$\begingroup$

Given the function, $y=f(x)=\frac3{2-x^2}$, find its domain and range.
The domain is of course = $R - \{-\sqrt2,\sqrt2\}$.
However, the range I got was wrong(rather incomplete).
Rewriting the function for x in terms of y, I got
$x=\pm \sqrt{\frac{2y-3}{y}} $
$\frac{2y-3}{y} \ge 0 \implies y\ge \frac32$
Therefore the $range(f) = [\frac32,\infty)$
However, the correct answer is $range(f)= [\frac32,\infty) \cup (-\infty,0)$
I dont understand why and how?

$\endgroup$
  • 3
    $\begingroup$ For $\lvert x\rvert > \sqrt{2}$, $f(x)$ is negative. Also, $\frac32$ is in the range, so $(-\infty,0)\cup [\frac32,\infty)$. $\endgroup$ – Daniel Fischer Nov 29 '13 at 12:05
  • $\begingroup$ @DanielFischer Damn it, that just skipped my mind! $\endgroup$ – Shaurya Gupta Nov 29 '13 at 12:07
3
$\begingroup$

Your mistake is that $$\frac{2y-3}{y} \geq 0 \Leftrightarrow (y>0 \wedge y \geq \frac32) \vee (y<0 \wedge y \leq \frac32) \Leftrightarrow y \in [\frac32, \infty) \cup (-\infty, 0)$$ You fogot the $y<0$ case, in which the relation sign reverses when mulltiplying by $y$.

$\endgroup$
  • $\begingroup$ what is the caret and reverse caret sign for? And why did you write $y\le3$ $\endgroup$ – Shaurya Gupta Nov 29 '13 at 12:11
  • 1
    $\begingroup$ $\wedge$: and, $\vee$: or $\endgroup$ – Kevin Nov 29 '13 at 12:15
  • 1
    $\begingroup$ @shauryagupta Assuming $y < 0$ we have $$\frac{2y-3}y \geq 0 \Rightarrow 2y - 3 \leq 0 \Rightarrow y \leq \frac32$$ So I made a little mistake in my head, thank you for bringing that to my attention :) $\endgroup$ – AlexR Nov 29 '13 at 12:20
  • 1
    $\begingroup$ And that's why I love StackExchange; aint never a doubt in my mind! $\endgroup$ – Shaurya Gupta Nov 29 '13 at 12:21
0
$\begingroup$

Since, the domain of $f(x)$ is $\mathbb{R}- \lbrace -2, 2 \rbrace$, so transform the equation into $x$ in terms of $y$, we will get $$ x= \sqrt{2 - \frac{3}{y}} $$

Now finding the domain of $x$ in terms of $y$ i.e., $$ \left( \frac{2y -3}{y} \right) \ge 0 $$ $$ \Rightarrow y \neq 0 $$ and $$ \Rightarrow y(2y - 3) \ge 0 $$ $$ \Rightarrow y \left(y - \frac {3}{2} \right)\ge 0 $$ $$ \Rightarrow y \in \left( -\infty ,0 \right) \cup \left ( \frac{3}{2}, \infty \right] $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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