Given, f(x) =\(\frac{x^2-9}{x-3}\)
Domain: Clearly, f(x) is defined for all x ∈ R expect x = 3
∴ Domain of f = R – {3} = (−∞, 3),∪ (3, ∞)
Range: Let y = f(x)
∴ y = \(\frac{x^2-9}{x-3}\)⇒ = + 3
It follow from the above relation that y takes all real values except 6 when takes values in the set
R – {3}
∴ Range of f = R – {6}