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?
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Shaurya Gupta
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3For $\lvert x\rvert > \sqrt{2}$, $f(x)$ is negative. Also, $\frac32$ is in the range, so $(-\infty,0)\cup [\frac32,\infty)$. – Daniel Fischer Nov 29 '13 at 12:05
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@DanielFischer Damn it, that just skipped my mind! – Shaurya Gupta Nov 29 '13 at 12:07
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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 $y$.
AlexR
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what is the caret and reverse caret sign for? And why did you write $y\le3$ – Shaurya Gupta Nov 29 '13 at 12:11
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1@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 :) – AlexR Nov 29 '13 at 12:20
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1And that's why I love StackExchange; aint never a doubt in my mind! – Shaurya Gupta Nov 29 '13 at 12:21
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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 $$ \Rightarrow y(2y - 3) \ge 0 $$ $$ \Rightarrow y \left(y - \frac {3}{2} \right)\ge 0 $$ $$ \Rightarrow y \in \left( -\infty ,0 \right) \cup \left ( \frac{3}{2}, \infty \right] $$
