Question:

Show that f :[−1,1]→R,given by f(x)= xx+2\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)= xx+2\frac {x}{x+2} , for some x in [−1, 1], i.e.,x= 2y1y\frac {2y}{1-y}

Updated On: Aug 24, 2023
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

f: [−1, 1] → R is given as  f(x)= xx+2\frac {x}{x+2} 
Let f(x) = f(y).
xx+2=yy+2\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= xx+2\frac {x}{x+2}
=>xy+2y=x
x(1-y)=2y  \Rightarrow x=2y1y\frac{2y}{1-y} ,y≠1
Now, let us define g: Range f → [−1, 1] as g(y)=2y1y\frac{2y}{1-y} ,y≠1
Now,(gof)(x)=g(f(x))=g (xx+2)(\frac {x}{x+2}) = 2(xx+2)1(xx+2)=2x2=x\frac{2(\frac {x}{x+2})}{1-(\frac{x}{x+2})}=\frac{2x}{2}=x.
Now,(gof)(x)=g(f(x))=g (2y1y)=2y1y2y1y+2=2y2=y.(\frac{2y}{1-y})= \frac{\frac{2y}{1-y}}{\frac{2y}{1-y+2}}=\frac{2y}{2}=y.
∴gof = I[-1,1] and fog = Irangef 
\therefore f−1 = g  \Rightarrow f-1 (y)=2y1y\frac{2y}{1-y} , y≠1

Was this answer helpful?
0
0

Top Questions on Relations and Functions

View More Questions