Question

Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer 1

f: R → R is given by,

f(x) = 4x + 3

One-one:

Let f(x) = f(y).

=> 4x + 3 = 4y + 3

=> 4x = 4y

=> x = y

∴ f is a one-one function.

Onto:

For y ∈ R, let y = 4x + 3.

`=> x = (y -3)/4 in R`

Therefore, for any y ∈ R, there exists `x= (y-3)/4 in R` such that

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

∴ f is onto.

Thus, f is one-one and onto and therefore, f⁻¹ exists.

Let us define g: R→ R by `g(x) = (y - 3)/4`  

Now, `(gof)(x) = g(f(x)) = g(4x + 3) = ((4x + 3) -3)/4 = x`

`(fog)(y) = f(g(y)) = f((y - 3)/4) =4((y-3)/4) + 3 = y-3+3 = y`

`∴gof = fog = I_R`

Hence, f is invertible and the inverse of f is given by

`f^(-1) = g(y) = (y-3)/4`