To find the domain of the function [tex]\( f(x) = \frac{1}{x^2 + 2x - 48} \)[/tex], we need to determine for which values of [tex]\( x \)[/tex] the function is defined. The function is undefined where the denominator is equal to zero because division by zero is undefined.
The denominator of the function is [tex]\( x^2 + 2x - 48 \)[/tex]. We need to find the values of [tex]\( x \)[/tex] for which this expression equals zero:
[tex]\[ x^2 + 2x - 48 = 0 \][/tex]
To find the roots of the quadratic equation, we solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 + 2x - 48 = 0 \][/tex]
Factoring the quadratic expression:
[tex]\[ (x + 8)(x - 6) = 0 \][/tex]
Setting each factor equal to zero gives us the critical points:
[tex]\[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \][/tex]
[tex]\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \][/tex]
So, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = -8 \)[/tex] and [tex]\( x = 6 \)[/tex]. These values must be excluded from the domain.
The domain of [tex]\( f(x) \)[/tex] will be all real numbers except [tex]\( -8 \)[/tex] and [tex]\( 6 \)[/tex]. In interval notation, this can be written as:
[tex]\[ (-\infty, -8) \cup (-8, 6) \cup (6, \infty) \][/tex]
Thus, the domain of the function [tex]\( f(x) = \frac{1}{x^2 + 2x - 48} \)[/tex] is:
[tex]\[ (-\infty, -8) \cup (-8, 6) \cup (6, \infty) \][/tex]
