|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

x+3x+3,x+3x+31,f(x)=2x+3x+32. Besides, f(x)0. Hence, f(R)[0,2]. Let us now show that [0,2]f(R). This implies that for every y[0,2] we need to find xR such that y=f(x). In fact it is sufficient to find x[3,0] satisfying the above conditions. We claim that x=y+23(y2) is the required value. Let us check this claim. 1. x=y+23(y2) must be from [3,0]. Indeed, since y2,y20, we have that y+23(y2)0, that is, x0. On the other hand, y0,2y0,y2y2,y2(y+2),y+2y21,y+23(y2)3,x3. This means that x[3,0]. 2. If x[3,0], then f(x)=3x2(x+3). Routine check shows that f(y+23(y2))=y, as required