(a). y = 3x + 4
The given function y = 3x + 4 gives a real value by replacing the independent variable x with any real number. Therefore, domain set of the function y = 3x + 4 = Domain = .
Now y = 3x + 4 ⇒ x = (y-4)/3.
Assigning any real number value to y satisfies the relation, therefore, the range R of the function is the set of real numbers ‘R’.

(b). f(t) = t² + 5
The given function gives a real value if we replace the independent variable t by any real number. Therefore, domain of the function D ={t|t∈R}.
Similarly, for t equal to any real number (positive or negative), its square is positive. Moreover, the range value f(t) will be 5 more than the square of t. Hence range of the function is the set of real numbers

c. SA = f(r) = 4πr²
The function represents the surface area of a sphere which is a function of the radius of the sphere. Now r (radius) of a sphere can’t be negative or 0. So r must be a real positive number greater than 0. Therefore, domain = D ={r|r>0, r∈R}.
Similarly, for any +ve value r, f(r) is also a +ve real number. Therefore, Range = R = {f(r)| f(r)>0}.