Question

Show that f :[−1,1]→R,given by f(x)= \(\frac {x}{x+2}\) is one-one. Find the inverse of the function f :[−1,1] \(\to\) Range f.
(Hint: For y ∈Range f, y = f(x)= \(\frac {x}{x+2}\) , for some x in [−1, 1], i.e.,x= \(\frac {2y}{1-y}\)

Answer 1

f: [−1, 1] → R is given as  f(x)= \(\frac {x}{x+2}\) 
Let f(x) = f(y).
\(\Rightarrow\frac {x}{x+2}=\frac{y}{y+2}\)
\(\Rightarrow\) xy+2x=xy+2y =>2x=2y
\(\Rightarrow\) x=y
∴ f is a one-one function.
It is clear that f: [−1, 1] → Range f is onto.
∴ f: [−1, 1] → Range f is one-one and onto and therefore, the inverse of the function:
f: [−1, 1] → Range f exists.
Let g: Range f → [−1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since f: [−1, 1] → Range f is onto, we have:y=f(x) for same x∈[-1,1]
y= \(\frac {x}{x+2}\)
=>xy+2y=x
x(1-y)=2y  \(\Rightarrow\) x=\(\frac{2y}{1-y}\) ,y≠1
Now, let us define g: Range f → [−1, 1] as g(y)=\(\frac{2y}{1-y}\) ,y≠1
Now,(gof)(x)=g(f(x))=g \((\frac {x}{x+2})\) = \(\frac{2(\frac {x}{x+2})}{1-(\frac{x}{x+2})}=\frac{2x}{2}=x\).
Now,(gof)(x)=g(f(x))=g \((\frac{2y}{1-y})= \frac{\frac{2y}{1-y}}{\frac{2y}{1-y+2}}=\frac{2y}{2}=y.\)
∴gof = I_[-1,1] and fog = I_rangef 
\(\therefore\) f⁻¹ = g  \(\Rightarrow\) f^-1 (y)=\(\frac{2y}{1-y}\) , y≠1