# Thread: Range and Domain of Functions

1. ## Range and Domain of Functions

Find the Domain of

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

Find the Domain of

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

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

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

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

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

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

Why is the 5/2 terms included?

2. 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:

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

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

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

so

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

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

Now what ???

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