Question

Let $f: R-\left\{-\frac{4}{3}\right\} \rightarrow R$ be a function defined as $f(x)=\frac{4 x}{3 x+4}$. The inverse of $f$ is the map $g:$ range $f \rightarrow R-\left\{-\frac{4}{3}\right\}$ given by

• A. $g(y)=\frac{3 y}{3-4 y}$
• B. $g(y)=\frac{4 y}{4-3 y}$
• C. $g(y)=\frac{4 y}{3-4 y}$
• D. $g(y)=\frac{3 y}{4-3 y}$

Answer 1

Answer: (B)

Given that, $f: R-\left\{-\frac{4}{3}\right\} \rightarrow R$ is defined as $f(x)=\frac{4 x}{3 x+4}$
Let $y$ be an arbitrary element of range of $f$.
Then, there exists $x \in R-\left\{-\frac{4}{3}\right\}$ such that $y=f(x)$
$\Rightarrow y=\frac{4 x}{3 x+4}$
$\Rightarrow 3 x y+4 y=4 x$
$\Rightarrow x(4-3 y)=4 y$
$\Rightarrow x=\frac{4 y}{4-3 y}$
Let us define $g$ : range $f \rightarrow R-\left\{-\frac{4}{3}\right\}$ as $g(y)=\frac{4 y}{4-3 y}$ :
Now, $(g o f)(x)=g(f(x))=g\left(\frac{4 x}{3 x+4}\right)$
$=\frac{4\left(\frac{4 x}{3 x+4}\right)}{4-3\left(\frac{4 x}{3 x+4}\right)}=\frac{16 x}{12 x+16-12 x}=\frac{16 x}{16}=x $
and $ (f o g)(y)=f(g(y))=f\left(\frac{4 y}{4-3 y}\right)$
$ =\frac{4\left(\frac{4 y}{4-3 y}\right)}{3\left(\frac{4 y}{4-3 y}\right)+4}=\frac{16 y}{12 y+16-12 y}$
$ =\frac{16 y}{16}=y $
Therefore, $g \circ f=I_{R-\left\{-\frac{4}{3}\right\}}$ and fog $=I_{\text {Range } f }$
Thus, $g$ is the inverse of $f$ i.e., $f^{-1}=g$.
Hence, the inverse of $f$ is the map $g:$ range $f \rightarrow R-\left\{-\frac{4}{3}\right\}$, which is given by $g(y)=\frac{4 y}{4-3 y}$.