To find the domain of the function [tex]\( f(x) = \frac{21}{x^2 + 2x - 168} \)[/tex], we need to determine the values of [tex]\( x \)[/tex] for which the function is defined. The function [tex]\( f(x) \)[/tex] is defined as long as the denominator is not zero. Therefore, we need to identify the values of [tex]\( x \)[/tex] that make the denominator equal to zero and exclude them from the domain.
The denominator of the function is [tex]\( x^2 + 2x - 168 \)[/tex]. To find the points of discontinuity, we set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
x^2 + 2x - 168 = 0
\][/tex]
This is a quadratic equation. To solve it, we can use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
In this case, the coefficients are [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -168 \)[/tex]. Substituting these values into the quadratic formula gives:
[tex]\[
x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-168)}}{2 \cdot 1}
= \frac{-2 \pm \sqrt{4 + 672}}{2}
= \frac{-2 \pm \sqrt{676}}{2}
= \frac{-2 \pm 26}{2}
\][/tex]
This results in two solutions:
[tex]\[
x = \frac{-2 + 26}{2} = \frac{24}{2} = 12
\][/tex]
[tex]\[
x = \frac{-2 - 26}{2} = \frac{-28}{2} = -14
\][/tex]
So, the quadratic equation [tex]\( x^2 + 2x - 168 = 0 \)[/tex] has roots at [tex]\( x = 12 \)[/tex] and [tex]\( x = -14 \)[/tex]. These are the values that make the denominator zero, and thus are the points of discontinuity for the function.
The domain of the function [tex]\( f(x) \)[/tex] will therefore be all real numbers except [tex]\( x = 12 \)[/tex] and [tex]\( x = -14 \)[/tex]. In interval notation, this is written as:
[tex]\[
\boxed{(-\infty, -14) \cup (-14, 12) \cup (12, \infty)}
\][/tex]
