Questions: Find the domain and range of the function f(x) = sqrt((x^2 + 3x - 18)/(x - 3)). Express with interval notation, using U for the union of sets and inf for infinity. Domain = Range =

Solution Steps

To find the domain and range of the function \( f(x) = \sqrt{\frac{x^2 + 3x - 18}{x - 3}} \):

1. Domain: Determine the values of \( x \) for which the expression inside the square root is non-negative and the denominator is not zero.
2. Range: Analyze the behavior of the function to determine the possible output values.

The function given is \( f(x) = \sqrt{\frac{x^2 + 3x - 18}{x - 3}} \).

To find the domain, we need to ensure that the expression inside the square root is non-negative and the denominator is not zero.

1. The denominator \( x - 3 \neq 0 \), so \( x \neq 3 \).
2. The expression inside the square root \( \frac{x^2 + 3x - 18}{x - 3} \geq 0 \).

Factor the quadratic expression in the numerator: \[ x^2 + 3x - 18 = (x + 6)(x - 3) \]

So the function becomes: \[ f(x) = \sqrt{\frac{(x + 6)(x - 3)}{x - 3}} \]

For \( x \neq 3 \), we can simplify the expression: \[ f(x) = \sqrt{x + 6} \]

The expression \( \sqrt{x + 6} \) is defined when \( x + 6 \geq 0 \): \[ x + 6 \geq 0 \] \[ x \geq -6 \]

Combining this with the restriction \( x \neq 3 \), the domain is: \[ x \in [-6, 3) \cup (3, \infty) \]

The function \( f(x) = \sqrt{x + 6} \) is a square root function, which outputs non-negative values. The minimum value of \( x + 6 \) is 0 (when \( x = -6 \)), and as \( x \) increases, \( x + 6 \) increases without bound.

Thus, the range of \( f(x) \) is: \[ [0, \infty) \]

Final Answer