The domain of the function in interval notation is \( (-\infty, 5) \cup(5, \infty) \). Part 2 of 5 (b) Simplify the rational expression defining the function. \( f(x)=\frac{x^{2}+x-30}{x-5} \) simplifies to \( f(x)=\square \) where \( x \neq \square \otimes \)

To simplify the expression \( f(x) = \frac{x^{2}+x-30}{x-5} \), we start by factoring the numerator. The expression \( x^2 + x - 30 \) can be factored into \( (x - 5)(x + 6) \). This gives us: \[ f(x) = \frac{(x - 5)(x + 6)}{x - 5} \] As long as \( x \neq 5 \), we can cancel \( (x - 5) \) from the numerator and denominator, yielding: \[ f(x) = x + 6 \quad \text{for} \quad x \neq 5 \] So, \( f(x) = x + 6 \) where \( x \neq 5 \).