Question

If f : R → R, such that f(x) = ax + b, a ≠ 0, then prove that f is a bijection function. Also, find f^-1.

Answer 1

Given function f : R→ R and ax + b, a ≠ 0

f^-1 exists if f : R → R be a bijection function, before this we have to prove f be a bijection function.

One-one/many-one:

Let p, q ∈ R

f(p) = f(q)

⇒ ap + b = aq + b

⇒ ap = aq

⇒ p = q

So, f(p) = f(q), ∀ p, q ∈ R

∴ f is a bijection function.

Onto/into:

Let f(x) = y, y ∈ R

ax + b = y

⇒ x = (y - b)/a∈ R

So, pre-image of every value of y exist in domain R,

∴ ‘f’ is onto function.

Thus, range of f = co-domain of f.

So, f is a bijection function

So, f^-1 exists.

Let, y ∈ R and f^-1(y)= x,

then f(x) = y

⇒ ax +b = y