Marylie,

Let me do a slightly different question

What are the domain and range of f(x) = √(x⁴ - 81)?

Domain:

It's the square root here that can cause a problem. You can't take the square root of a negative number so you must have

x⁴ - 81 ≥ 0

This means that

x⁴ ≥ 81

The fourth root of 81 is 3 so either x ≥ 3 or x ≤ -3. Hence the domain of f(x) is all real numbers x so that x ≥ 3 or x ≤ -3.

Range:

The question here is for what numbers y is it possible to find a number x so that y = √(x⁴ - 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 = √(x⁴ - 81)?

If you square both sides you get y² = x⁴ - 81 and hence x⁴ = y² + 81. The right side is positive so I can take the fourth root to get x = (y² + 81)^1/4.

Check:

You give me a non-negative number y and I take x = (y² + 81)^1/4. Then

√(x⁴ - 81) = √)[(y² + 81)^1/4]⁴ - 81) = √(y² + 81 - 81) = √y² and since y was non-negative √y² = y.

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

I hope this helps,

Harley