Solveeit Logo
Login

Question

Question: Let \(f:\left( 0,1 \right)\to \mathbb{R}\) be defined by \(f\left( x \right)=\dfrac{b-x}{1-bx}\), wh...

Let f:(0,1)→Rf:\left( 0,1 \right)\to \mathbb{R} be defined by f(x)=b−x1−bxf\left( x \right)=\dfrac{b-x}{1-bx}, where b is constant such that 0<b<10< b< 1, then
A. ff is not invertible on (0,1)\left( 0,1 \right)
B. f≠f−1f\ne {{f}^{-1}} on (0,1)\left( 0,1 \right) and f′(b)=−1f′(0){{f}^{'}}\left( b \right)=-\dfrac{1}{{{f}^{'}}\left( 0 \right)}
C. f=f−1f={{f}^{-1}} on (0,1)\left( 0,1 \right) and f′(b)=1f′(0){{f}^{'}}\left( b \right)=\dfrac{1}{{{f}^{'}}\left( 0 \right)}
D. f−1{{f}^{-1}} is invertible on (0,1)\left( 0,1 \right)

Explanation

Solution

We first try to find the inverse function for f(x)=b−x1−bxf\left( x \right)=\dfrac{b-x}{1-bx}. The domain for the inverse function would be the same as the range of the main function f(x)=b−x1−bxf\left( x \right)=\dfrac{b-x}{1-bx}. Then we try to point out the parts in the domain which are not well defined for the inverse function.

Complete step by step solution:
It is given that f:(0,1)→Rf:\left( 0,1 \right)\to \mathbb{R} be defined by f(x)=b−x1−bxf\left( x \right)=\dfrac{b-x}{1-bx}.
We first try to find the inverse of the given function.
We take y=f(x)=b−x1−bxy=f\left( x \right)=\dfrac{b-x}{1-bx}.
We try to find the value of xx based on yy.
Therefore,
y=b−x1−bx ⇒y−ybx=b−x ⇒x(1−by)=b−y ⇒x=b−y1−by \begin{aligned} & y=\dfrac{b-x}{1-bx} \\\ & \Rightarrow y-ybx=b-x \\\ & \Rightarrow x\left( 1-by \right)=b-y \\\ & \Rightarrow x=\dfrac{b-y}{1-by} \\\ \end{aligned}
Therefore, the inverse of the function ff is f−1=b−y1−by{{f}^{-1}}=\dfrac{b-y}{1-by}.
We have that the domain for ff is (0,1)\left( 0,1 \right) which becomes the range of the function f−1{{f}^{-1}} and the domain of f−1{{f}^{-1}} is R\mathbb{R}.
Now in f−1=b−y1−by{{f}^{-1}}=\dfrac{b-y}{1-by}, the denominator can’t be 0.
We can see for the value of y=1by=\dfrac{1}{b}, the value of 1−by1-by becomes 0.
We have 0<b<10< b< 1 which means y=1b∈(1,∞)y=\dfrac{1}{b}\in \left( 1,\infty \right).
So, in the domain of f−1=b−y1−by{{f}^{-1}}=\dfrac{b-y}{1-by}, there are parts which cannot define the inverse formula.
Therefore, ff is not invertible on (0,1)\left( 0,1 \right). The correct option is A.

Note: Now we find the differentiation of the f(x)=b−x1−bxf\left( x \right)=\dfrac{b-x}{1-bx}.
We get f′(x)=−(1−bx)+b(b−x)(1−bx)2=b2−1(1−bx)2{{f}^{'}}\left( x \right)=\dfrac{-\left( 1-bx \right)+b\left( b-x \right)}{{{\left( 1-bx \right)}^{2}}}=\dfrac{{{b}^{2}}-1}{{{\left( 1-bx \right)}^{2}}}. This gives f′(b)=b2−1(1−bx)2=1b2−1{{f}^{'}}\left( b \right)=\dfrac{{{b}^{2}}-1}{{{\left( 1-bx \right)}^{2}}}=\dfrac{1}{{{b}^{2}}-1}
We also have f′(0)=b2−1(1−bx)2=b2−1{{f}^{'}}\left( 0 \right)=\dfrac{{{b}^{2}}-1}{{{\left( 1-bx \right)}^{2}}}={{b}^{2}}-1
This means f′(b)=1f′(0){{f}^{'}}\left( b \right)=\dfrac{1}{{{f}^{'}}\left( 0 \right)}.