Let f - R -4-3- R be a function defined as f x -4 x -3 x -4- y -4-3- The inverse of f is the map g- Range f - R -4-3 is given bya gy-frac3 y3-4 y lb g y - frac 4 y4-3 yc gy- frac 4 y3-4 y d g y - frac 3 y 4-3 y l

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Definition of Relations

Let f : R --4...

Question

Let $f : R - {- \frac{4}{3}} \to R$ be a function defined as $f (x) = \frac{4 x}{3 x + 4}, x \neq - \frac{4}{3}$ . The inverse of f is the map g: Range $f \to R - {- \frac{4}{3}}$ is given by
$(a) g (y) = \frac{3 y}{3 - 4 y} (b) g (y) = \frac{4 y}{4 - 3 y} (c) g (y) = \frac{4 y}{3 - 4 y} (d) g (y) = \frac{3 y}{4 - 3 y}$

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Solution

Given that $f : R - {- \frac{4}{3}} \to R$ is defined as $f (x) = \frac{4 x}{3 x + 4}$
Let y be an arbitrary element of range f.
Then, there exists $x \in R - {- \frac{4}{3}}$ 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 $f \to R - {- \frac{4}{3}}$ as $g (Y) = \frac{4 y}{4 - 3 y} .$
Now, $(g o f) (x) = g (f (x)) = g (\frac{4 x}{3 x + 4})$
$= \frac{4 (\frac{4 x}{3 x + 4})}{4 - 3 (\frac{4 x}{3 x + 4})} = \frac{16 x}{12 x + 16 - 12 x} = \frac{16 x}{16} = x$
and $(f o g) (y) = f (g (y)) = f (\frac{4 y}{4 - 3 y})$
$= \frac{4 (\frac{4 y}{4 - 3 y})}{3 (\frac{4 y}{4 - 3 y}) + 4} = \frac{16 y}{12 y + 16 - 12 y} = \frac{16 y}{16} = y$
Therefore, $g o f = I_{R - {- \frac{4}{3}}}$ and $f o g = I_{R}$
Thus, g is the inverse of f i.e., $f^{- 1} = g$
Hence, the inverse of f is the map g:Range $f \to R - {- \frac{4}{3}}$ , which is given by $g (y) = \frac{4 y}{4 - 3 y}$ . Thus, the correct answer is (b).

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Q. Let

f : R - {- \frac{4}{3}} \to R

be a function defined as

f (x) = \frac{4 x}{3 x + 4}

. The inverse of

f

is the map

g : R a n g e f \to R - {- \frac{4}{3}}

given by

Q. If

f : R - {- \frac{4}{3}} \to R

be a function defined by

f (x) = \frac{4 x}{3 x + 4}

Show that the function

f

is invertible, Also find the inverse of

f

Consider $f : R - {- \frac{4}{3}} \to R - {\frac{4}{3}}$ given by $f (x) = \frac{4 x + 3}{3 x + 4}$ .Show that f is bijective.Find the inverse of f and Hence find $f^{- 1} (0)$ and x such that $f^{- 1} (x) = 2$ .

Q. Let

f : R \to R

be a function defined by

f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}

then

f

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Let f:R−{−43}→R be a function defined as f(x)=4x3x+4,x≠−43. The inverse of f is the map g: Range f→R−{−43} is given by (a)g(y)=3y3−4y(b)g(y)=4y4−3y(c)g(y)=4y3−4y(d)g(y)=3y4−3y