Let be a function defined as . The inverse of f is map g : Range (A) (B) (C) (D)

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Solution

The provided function is f( x )= 4x 3x+4 under the domain f:R−{ − 4 3 }→R .

Consider y as an arbitrary element of range of the function f .

y= 4x 3x+4 x( 4−3y )=4y x= 4y 4−3y g( y )= 4y 4−3y

The inverse of the function be g( y )= 4y 4−3y in the range f→R−{ −4 3 }

gof( x )=g( f( x ) ) =g( 4x 3x+4 ) = 4( 4x 3x+4 ) 4−3( 4x 3x+4 ) = 16x 12x+16−12x = 16x 16 =x

fog( x )=f( g( x ) ) =f( 4y 4−3y ) = 4( 4x 3x+4 ) 4−3( 4x 3x+4 ) = 16y 12y+16−12y = 16y 16 =y

The inverse of the provided function is,

f −1 ( y )=g( y ) = 4y 4−3y

Thus, the option (B) is the correct option.

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Similar questions

Letbe a function defined as. The inverse of f is map g: Range

(A) (B)

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

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
$(a) g (y) = \frac{3 y}{3 - 4 y} (b) g (y) = \frac{4 y}{4 - 3 y} (c) g (y) = \frac{4 y}{3 - 4 y} (d) g (y) = \frac{3 y}{4 - 3 y}$