# Range and Domain of Functions

• Mar 4th 2011, 10:56 AM
cupid
Range and Domain of Functions
Find the Domain of
http://www.mathhelpforum.com/math-he...tions-q-3-.jpg

I can do this Question using graphs. I need help in doing it without graphs ... some help please

Find the Domain of
http://www.mathhelpforum.com/math-he...nctions-28.jpg

$-1 <= [2x^2 - 3] <= 1$

$-1 <= (2x^2 - 3) < 2$

$1 <= x^2 < \frac{5}{2}$

Why?
$|x| >= 1$ and $|x| < \sqrt{\frac{5}{2}}$

$x \in (-\sqrt{\frac{5}{2}},-1]\cup[1,\sqrt{\frac{5}{2}})$

$x \in [-\sqrt{\frac{5}{2}},-1]\cup[1,\sqrt{\frac{5}{2}}]$

Why is the 5/2 terms included?
• Mar 5th 2011, 12:49 PM
cupid
never mind about the second one ... its probably editor's carelessness

But i really need help with the first one

I've got this so far:

$|sin^{-1}sin x| \geq cos^{-1}cos x$

$|sin^{-1}sin x| \ = \ sin^{-1}sin x, \ when \ sin^{-1}sin x \geq 0 \implies sin x\geq0 \implies x \in [0,\pi] \ in \ [0,2\pi]$

$= \ -sin^{-1}sin x \ when \ sin^{-1}sin x \leq 0 \implies sin x \leq 0 \implies x \in [\pi,2\pi] \ in \ [0,2\pi]$

so

$sin^{-1}sin x \geq cos^{-1}cos x \ \forall \ x \in [0,\pi]$

$-sin^{-1}sin x \geq cos^{-1}cos x \ \forall \ x \in [\pi,2\pi]$

Now what ???

Please Reply ... i have nothing new to post now ... and the moderator will kill me if i again dump in a thread