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

विकल्प

• `g(y) = (3y)/(3 - 4y)`

• `g(y) = (4y)/(4 - 3y)`

• `g(y) = (4y)/(3 - 4y)` 

• `g(y) = (3y)/(4 - 3y)`

Answer 1

`bb(g(y) = (4y)/(4 - 3y))`

Explanation:

It is given that `f: R -{-4/3} → R` is defined as `f(x) = (4x)/(3x - 4)`.

Let y be an arbitrary element of Range f.

Then, there exists `x ∈ R - {-4/3}` such that y = f(x).

⇒ `y = (4x)/(3x +  4)`

⇒ 3xy + 4y = 4x 

⇒ x(4 – 3y) = 4y

⇒ `x = (4y)/(4 - 3y)`

Let us define g: Range `f → R -{-4/3}` as `g(y) = (4y)/(4 - 3y)`.

Now, (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

And (fog)(y) = f(g(y)) 

= `f((4y)/(4 - 3y))`

= `(4((4y)/(4 - 3y)))/(3((4y)/(4 - 3y)) + 4)`

= `(16y)/(12y + 16 - 12y)`

= `(16y)/16`

= y

∴ `gof = I_(R - {-4/3})` and `fog = I_"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 → R - {-4/3}`, which is given by `g(y) = (4y)/(4 - 3y)`.