f(x) = x2, g (x) = 5x – 6, Domain = (2, 3, 4} For fix) =x2, Range Ry= {f(2), f(3), f{4}}
= (4, 9, 16}
For g (x) = 5x – 6,
Range Rf = {g (2), g (3), g (4)}
∴ g(2) = 5 (2) – 6 = 10 – 6 = 4 ,
g (3) = 5(3) – 6 = 15 – 6 = 9
g (4) = 5 (4) – 6 = 20 – 6 = 14
Rf = {4, 9, 14}
Domains of the functions fix) and g (x) are same but their range are not same. Hence, the given functions are unequal functions.