Let f : R– {-(1/3)} → R be a function defined as f(x) = 4x/(3x + 4) . Show that, in f : R– {-(4/3)} → Range of f, f is one-one and onto. ​

Question

Let f : R– {\(-\frac{1}{3}\)} → R be a function defined as f(x) = \(\frac{4x}{3x+4}\). Show that, in f : R– {\(-\frac{4}{3}\)} → Range of f, f is one-one and onto. Hence find f⁻¹ : Range f→ R– {\(-\frac{4}{3}\)}.

Answer 1

Let x₁, x₂ ∈ R - {\(-\frac{4}{3}\)}

Now, 

f(x₁) = f(x₂)

⇒ \(\frac{4x_1}{3x_1+4}=\frac{4x_2}{3x_2+4}\)

⇒ 12 x₁ x₂ + 16 x₁ = 12 x₁ x₂ + 16 x₂

⇒ 16 x₁ = 16x₂

⇒ x₁ = x₂

Hence f is one-one function

Since, co-domain f is range of f

So, f : |R - {\(-\frac{4}{3}\)} → |R in one-one onto function.

For inverse function,

Let f(x) = y

Therefore, f^-1 :  Range of f → R-{-(4/3)} is 

f^-1(y) = \(\frac{4y}{4-3y}\)