To determine the domain of the rational expression
[tex]\[ f(x) = \frac{-1}{x^2 + 2x - 35}, \][/tex]
we must identify the values of [tex]\( x \)[/tex] that make the denominator equal to zero, since the expression is undefined for those values.
Firstly, we need to factor the quadratic expression in the denominator:
[tex]\[ x^2 + 2x - 35. \][/tex]
We look for two numbers that multiply to [tex]\(-35\)[/tex] and add to [tex]\(2\)[/tex]. These numbers are [tex]\(7\)[/tex] and [tex]\(-5\)[/tex]:
[tex]\[ x^2 + 2x - 35 = (x + 7)(x - 5). \][/tex]
So, the denominator can be rewritten as:
[tex]\[ x^2 + 2x - 35 = (x + 7)(x - 5). \][/tex]
Next, we set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ (x + 7)(x - 5) = 0. \][/tex]
This gives us two solutions:
[tex]\[ x + 7 = 0 \quad \Rightarrow \quad x = -7, \][/tex]
[tex]\[ x - 5 = 0 \quad \Rightarrow \quad x = 5. \][/tex]
Therefore, the rational expression is undefined at [tex]\( x = -7 \)[/tex] and [tex]\( x = 5 \)[/tex]. These values are not in the domain of the function.
The domain of the function is all real numbers except [tex]\( x = -7 \)[/tex] and [tex]\( x = 5 \)[/tex]. In interval notation, this is written as:
[tex]\[ (-\infty, -7) \cup (-7, 5) \cup (5, \infty). \][/tex]
Thus, the domain is:
[tex]\[
\boxed{(-\infty, -7) \cup (-7, 5) \cup (5, \infty)}
\][/tex]
