Question

SOLUTION We have
[
begin{aligned}
fleft(x_{1}right)=fleft(x_{2}right) & Rightarrow frac{x_{1}}{x_{1}+2}=frac{x_{2}}{x_{2}+2}
& Rightarrow x_{1} x_{2}+2 x_{1}=x_{1} x_{2}+2 x_{2}
& Rightarrow 2left(x_{1}-x_{2}right)=0
& Rightarrow x_{1}-x_{2}=0
& Rightarrow x_{1}=x_{2}
end{aligned}
]
( therefore quad f ) is one-one.
since range ( (f)=Y, ) so ( f ) is onto. Thus, ( f ) is one-one onto and therefore invertible. Let ( y in Y ). Then, there exists ( x in[-1,1] ) such that ( f(x)=y . )
[
begin{aligned}
text { Now, } y=f(x) & Rightarrow y=frac{x}{(x+2)}
& Rightarrow x=frac{2 y}{(1-y)}
& Rightarrow f^{-1}(y)=frac{2 y}{(1-y)}
end{aligned}
]
Thus, we define:
[
f^{-1}:[-1,1] rightarrow Y: f^{-1}(y)=frac{2 y}{(1-y)}, y eq 1
]

# Let f:(-1, 1]=> Y:f(x) = 12, x +-2 and Y = range (f ). Show that f is invertible and find f-1. (x + 2)

Solution