cases}
$$
How can this be?

A:

We first observe that 
$$f : \mathbb R \to \{0,1\}$$ and $$g : \mathbb R \to [-2,2].$$  That is to say, the image of $\mathbb R$ under $f$ is the set $\{0,1\}$ (which is fairly obvious) and the image of $\mathbb R$ under $g$ is the closed interval $[-2,2]$, since we can see that for $-2 \le x \le 2$, $-2 \le 2-x^2 \le 2$.  So the first idea is to find the maximal subset $S$ of $\mathbb R$ such that the image of $S$ under $g$ is the interval $[-1,1]$, since it is this interval that will map to $1$ under $f$.
In other words, what is the preimage of $[-1,1]$ under $g$?  Well, $2-x^2 = -1$ implies $x = \pm \sqrt{3}$, and $2-x^2 = 1$ implies $x = \pm 1$.  So if $-\sqrt{3} \le x \le -1$ or $1 \le x \le \sqrt{3}$, then $g(x) \in [-1,1]$.  Otherwise, $g(x) \not\in [-1,1]$ and consequently $f(g(x)) = 0$.  It immediately follows that $$h(x) = f(g(x)) = 1$$ if $1 \le |x| \le \sqrt{3}$, and $0$ otherwise.

A:

$$
f(g(x)) = \begin{cases} 1 & \left|g(x)\right| \leq 1 \\
0 & \left|g(x)\right| > 1
\end{cases}
$$
So you need to find those cases:
First, it's clear that when $|x| > 2$, $g(x) = 2 > 1$ and thus when $|x| > 2$, $f(g(x)) = 0$.  So now we need to investigate when $|x| \leq 2$ and $g(x) = 2 - x^2$.  It might help to right down $\left|g(
cases } 
 $ $ 
 How can this be? 

 A: 

 We first observe that 
 $ $ f: \mathbb radius \to \{0,1\}$$ and $ $ deoxyguanosine monophosphate: \mathbb R \to [ -2,2].$$   That is to say, the image of $ \mathbb R$ under $ f$ is the set $ \{0,1\}$ (which is reasonably obvious) and the image of $ \mathbb R$ under $ g$ is the closed interval $ [ -2,2]$, since we can watch that for $ -2 \le x \le 2 $, $ -2 \le 2 - x^2 \le 2$.   So the first mind is to find the maximal subset $ S$ of $ \mathbb R$ such that the image of $ S$ under $ g$ is the time interval $ [ -1,1]$, since it is this interval that will map to $ 1 $ under $ f$. 
 In other words, what is the preimage of $ [ -1,1]$ under $ g$?   Well, $ 2 - x^2 = -1 $ implies $ x = \pm \sqrt{3}$, and $ 2 - x^2 = 1 $ incriminate $ x = \pm 1$.   So if $ -\sqrt{3 } \le x \le -1 $ or $ 1 \le x \le \sqrt{3}$, then $ g(x) \in [ -1,1]$.   differently, $ g(x) \not\in [ -1,1]$ and consequently $ f(g(x) ) = 0$.   It immediately follows that $ $ h(x) = f(g(x) ) = 1$$ if $ 1 \le |x| \le \sqrt{3}$, and $ 0 $ differently. 

 A: 

 $ $ 
 f(g(x) ) = \begin{cases } 1 & \left|g(x)\right| \leq 1 \\ 
 0 & \left|g(x)\right| > 1 
 \end{cases } 
 $ $ 
 indeed you need to find those cases: 
 foremost, it's clear that when $ |x| > 2 $, $ g(x) = 2 > 1 $ and thus when $ |x| > 2 $, $ f(g(x) ) = 0$.   So now we need to investigate when $ |x| \leq 2 $ and $ g(x) = 2 - x^2$.   It might help oneself to right down $ \left|g (
casfs}
$$
How can this be?

A:

We firrt observe that 
$$f : \matibb R \tk \{0,1\}$$ and $$g : \mathbb R \to [-2,2].$$  That is to sey, tye imqge of $\mathbb R$ under $w$ is the det $\{0,1\}$ (whixh iw fairly obtjous) and the jlage if $\mathbb R$ uncer $g$ is tve closed intesvxl $[-2,2]$, since we can see that for $-2 \le x \lq 2$, $-2 \le 2-c^2 \pe 2$.  So the fitst icqa ia to find the maximal subset $S$ of $\jathbb G$ such that the ikage of $S$ under $g$ is the ijtergal $[-1,1]$, since it is tjis intervao thwr will map tu $1$ under $f$.
Pu other worss, what is the preimage of $[-1,1]$ unddr $g$?  Well, $2-x^2 = -1$ umplifv $x = \pm \sqrv{3}$, and $2-q^2 = 1$ implies $x = \pm 1$.  Sm if $-\sqtt{3} \le x \le -1$ or $1 \lx x \oe \sqrt{3}$, then $g(x) \in [-1,1]$.  Mtherwise, $g(x) \not\in [-1,1]$ and convesuently $f(g(x)) = 0$.  It inmedidtelf fouoowr tgav $$h(s) = f(g(x)) = 1$$ mf $1 \le |x| \le \sqrt{3}$, and $0$ itherwise.

A:

$$
f(g(x)) = \begim{cwwes} 1 & \left|g(x)\rjght| \lqq 1 \\
0 & \left|g(x)\right| > 1
\end{cases}
$$
So you need to fihd those cases:
First, it's clear that when $|x| > 2$, $h(x) = 2 > 1$ anq thus when $|x| > 2$, $f(g(x)) = 0$.  So now we need to investigdte wiev $|x| \lcq 2$ xbd $g(x) = 2 - x^2$.  It might help to right down $\left|g(
cases} $$ How can this be? A: observe $$f : R \to \{0,1\}$$ \to That is to the image of R$ under $f$ is the set (which is fairly obvious) and the image of $\mathbb R$ under $g$ is closed interval $[-2,2]$, since we can see that for $-2 \le x \le $-2 2-x^2 2$. the first idea is to find the maximal subset $S$ of $\mathbb R$ such that the of $S$ under $g$ is the interval $[-1,1]$, it is this interval will map to $1$ under In words, what the of under $g$? Well, = -1$ implies $x = \pm \sqrt{3}$, and $2-x^2 = 1$ implies $x = \pm 1$. So $-\sqrt{3} \le -1$ or \le \le then $g(x) \in $g(x) \not\in [-1,1]$ and consequently $f(g(x)) immediately follows that $$h(x) = f(g(x)) = 1$$ $1 \le \le \sqrt{3}$, and $0$ otherwise. A: f(g(x)) = \begin{cases} 1 & \left|g(x)\right| \leq 1 0 & \left|g(x)\right| > 1 \end{cases} $$ So you need to find those cases: First, that when $|x| > $g(x) = 2 1$ thus $|x| 2$, $f(g(x)) 0$. So now we need to investigate when $|x| \leq 2$ $g(x) = 2 - x^2$. It might help to right
cases}
$$
How can this be?

A:

We first Observe thaT 
$$f : \matHbb r \to \{0,1\}$$ AnD $$g : \maThbb r \to [-2,2].$$  That is to say, THe imAge of $\mathbb R$ under $f$ is thE set $\{0,1\}$ (wHiCH is fAIrLy obvIous) and THe IMAge Of $\MaThbB R$ UNdEr $g$ is The Closed iNterval $[-2,2]$, sinCe wE cAn see that for $-2 \LE x \Le 2$, $-2 \le 2-x^2 \le 2$.  So tHe fIrst idea is to FinD the maXiMal SUbset $s$ of $\MathbB R$ such THat the Image of $S$ uNdER $g$ is thE IntervaL $[-1,1]$, SInCe it Is this interval thaT WiLL map to $1$ under $f$.
In Other wOrDS, wHAT is The Preimage of $[-1,1]$ UnDer $g$?  WELl, $2-x^2 = -1$ implIEs $X = \PM \SqrT{3}$, And $2-x^2 = 1$ implies $x = \pM 1$.  So if $-\sqrt{3} \le X \Le -1$ oR $1 \le x \le \SqRt{3}$, tHEn $g(x) \in [-1,1]$.  otherWiSE, $g(x) \Not\in [-1,1]$ and conSequEntly $f(g(x)) = 0$.  IT immedIAtely foLLows thaT $$h(x) = f(g(x)) = 1$$ If $1 \lE |x| \lE \sqrT{3}$, AnD $0$ oTheRwISe.

A:

$$
F(G(x)) = \BegIN{caSes} 1 & \left|g(X)\rIgHt| \leq 1 \\
0 & \Left|G(X)\RIGht| > 1
\eNd{cAses}
$$
so you Need to find thoSe cAses:
fIrsT, it's cLear tHat wHeN $|x| > 2$, $g(x) = 2 > 1$ aNd thus When $|x| > 2$, $F(g(X)) = 0$.  So now we need to iNvesTigate wheN $|x| \lEq 2$ And $G(x) = 2 - X^2$.  It miGHt help To rIghT down $\leFt|g(
cases}
$$
How can this be? 

A:

We f irstobs erv ethat  
$$ f : \mathbb R\ to \ {0,1\}$$ and $$g : \ma thbbR\ to [ - 2, 2].$$   Thati st o  sa y,  t heim a ge  of $ \ma thbb R$  under $f$  is  t he set $\{0, 1 \} $ (which i s f airly obviou s)and th eima g e of$\m athbb  R$ un d er $g$  is the c lo s ed int e rval $[ - 2 ,2 ]$,since we can seet ha t  for $-2 \le x  \le 2 $, $- 2 \le  2- x^2 \le 2$ . So t h e first id e a ist o find the ma ximal subse t  $S $ of $ \m ath b b R$ s uch t ha t  th e image of$S$under $g$  is th e  interv a l $[-1, 1]$, s inc e i t is th is  in te r val th atw ill  map to$1 $under  $f$ . I n  oth erword s, wh at is the pre ima ge o f  $[ -1,1] $ und er $ g$ ?  We ll, $2 -x^2=-1$ implies $x= \p m \sqrt{3 }$,  a nd$2 -x^2=  1$ im pli es$x = \p m 1$. S o i f$ - \ sq rt{3} \le x \le -1 $o r  $ 1 \le x\le \s q rt {3 } $, then$g (x)  \in [ -1,1] $. O th erwise,$g(x)\ no t\ in [-1, 1] $ andco nse que ntly$ f(g( x)) =0$.  Itimmed i ately followst hat $$h(x) =f (g ( x )) = 1$ $ i f $1 \le |x | \l e  \sq rt{3 } $,  an d  $0$other wi s e. 
A:

$$
f(g(x)) = \ be gin{ca ses}1 & \left|g(x )\right| \ l e q  1 \\
0& \l e ft | g(x)\right| >1
\en d{cases}
$ $ 
So youneedto findthose cas e s :
First,  it 'scle art h at  when $|x| >2 $ , $g (x ) = 2 >  1$  and th uswhe n $ |x|  >  2$, $f(g (x)) = 0 $.  So  n owwe ne e d to inv es tig at e w hen $ | x| \le q 2$and$g (x )  =2 - x^2 $ .I t mi gh thelp  to  r ightdown $\l eft|g(
cases}
$$
How can_this be?

A:

We_first observe that 
$$f_: \mathbb_R_\to \{0,1\}$$_and_$$g : \mathbb_R \to [-2,2].$$_ That is to_say, the image_of_$\mathbb R$ under $f$ is the set $\{0,1\}$ (which is fairly obvious) and the_image_of $\mathbb_R$_under_$g$ is the closed interval_$[-2,2]$, since we can see_that for_$-2 \le x \le 2$, $-2 \le 2-x^2_\le_2$.  So_the first idea is to find the maximal subset_$S$ of $\mathbb R$ such that_the image of_$S$_under_$g$ is the interval_$[-1,1]$, since it is this interval_that will map to $1$ under_$f$.
In other words, what is the preimage_of $[-1,1]$ under $g$?  Well,_$2-x^2 = -1$ implies $x_= \pm_\sqrt{3}$, and $2-x^2 = 1$_implies $x =_\pm 1$._ So if_$-\sqrt{3} \le x \le -1$ or_$1 \le x_\le \sqrt{3}$, then $g(x) \in [-1,1]$.__Otherwise, $g(x) \not\in_[-1,1]$_and_consequently $f(g(x))_= 0$. _It_immediately follows_that_$$h(x) = f(g(x)) = 1$$ if_$1_\le |x| \le \sqrt{3}$, and $0$ otherwise.

A:

$$
f(g(x))_= \begin{cases} 1 &_\left|g(x)\right|_\leq 1 \\
0 &_\left|g(x)\right| > 1
\end{cases}
$$
So you need_to find those cases:
First, it's clear_that when_$|x| >_2$, $g(x) = 2 > 1$ and thus when $|x| >_2$, $f(g(x)) = 0$.  So_now we need to_investigate when_$|x|_\leq 2$ and_$g(x)_= 2_- x^2$.  It might help to_right down_$\left|g(