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|x+3| Besides, f(x)>=0 . Hence, f(R)sube[0,2] . Let us now show that [0,2]subef(R) . This implies that for every yin[0,2] we need to find
|x+3|\ Besides,
f(x)>=0
. Hence,
f(R)sube[0,2]
.\ Let us now show that
[0,2]subef(R)
. This implies that for every
yin[0,2]
we need to find
xinR
such\ that
y=f(x)
. In fact it is sufficient to find
xin[-3,0]
satisfying the above conditions. We claim\ that
x=(3(y-2))/(y+2)
is the required value.\ Let us check this claim.\
x=(3(y-2))/(y+2)
must be from
-3,0
. Indeed, since
y, we have that
(3(y-2))/(y+2),\ that is,
x. On the other hand,\
y>=0,2y>=0,y-2>=-y-2,y-2>=-(y+2),(y-2)/(y+2)>=-1,(3(y-2))/(y+2)>=-3,\ x>=-3.
\ This means that
xin[-3,0]
.\ If
xin[-3,0]
, then
f(x)=(2(x+3))/(3-x)
. Routine check shows that
f((3(y-2))/(y+2))=y
, as required.\ \ \ \ I need help witht the explanation of this answer. i did not why he choosed the interval [-3,0] and why also x=(2(x+3))/(3-x)\ \ please explain
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