Question

Let f: R - { -4/3 } → R be a function defined as f(x) = 4x / (3x+4). The inverse of f is map g: Range
f → R - { - 4/3 } given by

(A) g(y) = 3y / (3 - 4y)
(B) g(y) = 4y / (4 - 3y)
(C) g(y) = 4y / (3 - 4y)
(D) g(y) = 3y / (4 - 3y)

Answer 1

$f:\mathbb{R}-\{-4/3\}\rightarrow\mathbb{R}$be a function defined as $f(x)=\frac{4x}{3x+4}$

A/C to question, the inverse of f is map $g:\mathbb{R}\rightarrow\mathbb{R}-\{-4/3\}$

in simple way we have to find out inverse of f

so, Let's find it ,
$f(x)=y=\frac{4x}{3x+4}\\\implies y(3x+4)=4x\\\implies 3xy + 4y = 4x \\\implies x = \frac{4y}{4-3y}$
hence, $f^{-1}(x)=\frac{4x}{4-3x}$
so, $g= f^{-1}$
hence,$g(y) = \frac{4y}{4-3y}$
hence, option (B) is correct.

Answer 2

Answer:

Step-by-step explanation:f:\mathbb{R}-\{-4/3\}\rightarrow\mathbb{R} be a function defined as f(x)=\frac{4x}{3x+4}

A/C to question, the inverse of f is map g:\mathbb{R}\rightarrow\mathbb{R}-\{-4/3\}

in simple way we have to find out inverse of f

so, Let's find it ,

f(x)=y=\frac{4x}{3x+4}\\\implies y(3x+4)=4x\\\implies 3xy + 4y = 4x \\\implies x = \frac{4y}{4-3y}

hence, f^{-1}(x)=\frac{4x}{4-3x}

so, g= f^{-1}

hence, g(y) = \frac{4y}{4-3y}