Can anyone tell me how my book went from point a to b?
$\displaystyle
|x^2 - 1| < 2 \Rightarrow - \sqrt 3 < x < \sqrt 3
$
I just don't see it...thanks.
Try drawing the graphs of $\displaystyle y = |x^2 - 1|$ and $\displaystyle y = 2$. Now calculate the x-coordinate of the intersection points of these two graphs:
$\displaystyle x^2 - 1 = 2 \Rightarrow x^2 = 3 \Rightarrow x = \sqrt{3}$ or $\displaystyle x = -\sqrt{3}$.
For what values of x is the graph of $\displaystyle y = |x^2 - 1|$ below the graph of $\displaystyle y = 2$ ....?
Hello,
Why doesn't it work here ?
$\displaystyle |x^2-1|<2 \Leftrightarrow -2<x^2-1<2 \Leftrightarrow -1<x^2<3$
Whatever x is, the latter inequality can be restricted to :
$\displaystyle x^2<3$, since $\displaystyle x^2 \geq 0$.
So we have this new inequality :
$\displaystyle x^2<3$
which gives $\displaystyle |x|< \sqrt{3}$ (*)
that is to say $\displaystyle -\sqrt{3}<x<\sqrt{3}$
~~~~~~~~~~~~~~~~~~~~~~
note : $\displaystyle |a|<b \Leftrightarrow -b<a<b$, where b is positive.
(*) this can be proved.
a>0, $\displaystyle x^2<a^2 \Leftrightarrow |x|<a$
$\displaystyle x^2-a^2<0$
$\displaystyle (x-a)(x+a)<0$
so either $\displaystyle x-a<0$ and $\displaystyle x+a>0$ ----> x < a and x > -a. This is equivalent to |x|< a
either $\displaystyle x-a>0$ and $\displaystyle x+a<0$ -----> x > a and x < -a. This is not possible since a > 0.