Answer :
To find all values of [tex]\( x \)[/tex] that are NOT in the domain of the function [tex]\( f(x) = \frac{x^2 + 11x + 30}{x^2 - 64} \)[/tex], we need to determine where the denominator is equal to zero. Division by zero is undefined, so these values need to be excluded from the domain.
The denominator of the function is [tex]\( x^2 - 64 \)[/tex]. We need to set this equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 64 = 0 \][/tex]
To solve this equation, we can factor it:
[tex]\[ (x - 8)(x + 8) = 0 \][/tex]
Now, we set each factor equal to zero:
[tex]\[ x - 8 = 0 \quad \text{or} \quad x + 8 = 0 \][/tex]
Solving these simple linear equations, we get:
[tex]\[ x = 8 \quad \text{or} \quad x = -8 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] that are NOT in the domain of the function [tex]\( f \)[/tex] are [tex]\( 8 \)[/tex] and [tex]\( -8 \)[/tex].
So, the values of [tex]\( x \)[/tex] that are NOT in the domain of [tex]\( f \)[/tex] are:
[tex]\[ x = 8, -8 \][/tex]
The denominator of the function is [tex]\( x^2 - 64 \)[/tex]. We need to set this equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 64 = 0 \][/tex]
To solve this equation, we can factor it:
[tex]\[ (x - 8)(x + 8) = 0 \][/tex]
Now, we set each factor equal to zero:
[tex]\[ x - 8 = 0 \quad \text{or} \quad x + 8 = 0 \][/tex]
Solving these simple linear equations, we get:
[tex]\[ x = 8 \quad \text{or} \quad x = -8 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] that are NOT in the domain of the function [tex]\( f \)[/tex] are [tex]\( 8 \)[/tex] and [tex]\( -8 \)[/tex].
So, the values of [tex]\( x \)[/tex] that are NOT in the domain of [tex]\( f \)[/tex] are:
[tex]\[ x = 8, -8 \][/tex]