The question here is for what numbers y is it possible to find a number x so that y = √(x^{4} - 81)? The operator √ always returns a non-negative number so y must be non-negative but given a non-negative number y can you find an x so that y = √(x^{4} - 81)?

If you square both sides you get y^{2} = x^{4} - 81 and hence x^{4} = y^{2} + 81. The right side is positive so I can take the fourth root to get x = (y^{2} + 81)^{1/4}.

Check:

You give me a non-negative number y and I take x = (y^{2} + 81)^{1/4}. Then

√(x^{4} - 81) = √)[(y^{2} + 81)^{1/4}]^{4} - 81) = √(y^{2} + 81 - 81) = √y^{2} and since y was non-negative √y^{2} = y.

Thus the range of f(x) is all non-negative numbers.