Question

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

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

Answer 1

Answer: (B)

Given that, $f:R-\left\{-\frac{4}{3}\right\}\to R$ is defined as $f(x)=\frac{4x}{3x+4}$
Let $y$ be an arbitrary element of range of $f$.
Then, there exists $x\in R-\left\{-\frac{4}{3}\right\}$ such that $y=f(x)$
$\Rightarrow y=\frac{4x}{3x+4}$
$\Rightarrow 3xy+4y=4x$
$\Rightarrow x(4-3y)=4y$
$\Rightarrow x=\frac{4y}{4-3y}$
Let us define $g$ : range $f\to R-\left\{-\frac{4}{3}\right\}$ as $g(y)=\frac{4y}{4-3y}$ :
Now, $(gof)(x)=g(f(x))=g\left(\frac{4x}{3x+4}\right)$
$=\frac{4\left(\frac{4x}{3x+4}\right)}{4-3\left(\frac{4x}{3x+4}\right)}=\frac{16x}{12x+16-12x}=\frac{16x}{16}=x$
and $(fog)(y)=f(g(y))=f\left(\frac{4y}{4-3y}\right)$
$=\frac{4\left(\frac{4y}{4-3y}\right)}{3\left(\frac{4y}{4-3y}\right)+4}=\frac{16y}{12y+16-12y}$
$=\frac{16y}{16}=y$
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 $f\to R-\left\{-\frac{4}{3}\right\}$, which is given by $g(y)=\frac{4y}{4-3y}$.