Answer :
To determine the domain of the function [tex]\( f(x) = \frac{7}{x+11} + \frac{9}{x+8} \)[/tex], we need to identify the values of [tex]\( x \)[/tex] for which the function is defined. The function is defined as long as the denominators of each fraction are not zero.
1. Consider the first fraction [tex]\(\frac{7}{x+11}\)[/tex]:
- The denominator [tex]\( x+11 \)[/tex] must not be zero.
- Solving [tex]\( x+11 = 0 \)[/tex], we find [tex]\( x = -11 \)[/tex].
2. Consider the second fraction [tex]\(\frac{9}{x+8}\)[/tex]:
- The denominator [tex]\( x+8 \)[/tex] must not be zero.
- Solving [tex]\( x+8 = 0 \)[/tex], we find [tex]\( x = -8 \)[/tex].
Therefore, the function [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = -11 \)[/tex] and [tex]\( x = -8 \)[/tex]. These are the points where division by zero would occur.
Hence, the domain of the function [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = -11 \)[/tex] and [tex]\( x = -8 \)[/tex].
In interval notation, this is represented as:
[tex]\[ (-\infty, -11) \cup (-11, -8) \cup (-8, \infty) \][/tex]
Thus, the domain of [tex]\( f(x) \)[/tex] is [tex]\( (-\infty, -11) \cup (-11, -8) \cup (-8, \infty) \)[/tex].
1. Consider the first fraction [tex]\(\frac{7}{x+11}\)[/tex]:
- The denominator [tex]\( x+11 \)[/tex] must not be zero.
- Solving [tex]\( x+11 = 0 \)[/tex], we find [tex]\( x = -11 \)[/tex].
2. Consider the second fraction [tex]\(\frac{9}{x+8}\)[/tex]:
- The denominator [tex]\( x+8 \)[/tex] must not be zero.
- Solving [tex]\( x+8 = 0 \)[/tex], we find [tex]\( x = -8 \)[/tex].
Therefore, the function [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = -11 \)[/tex] and [tex]\( x = -8 \)[/tex]. These are the points where division by zero would occur.
Hence, the domain of the function [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = -11 \)[/tex] and [tex]\( x = -8 \)[/tex].
In interval notation, this is represented as:
[tex]\[ (-\infty, -11) \cup (-11, -8) \cup (-8, \infty) \][/tex]
Thus, the domain of [tex]\( f(x) \)[/tex] is [tex]\( (-\infty, -11) \cup (-11, -8) \cup (-8, \infty) \)[/tex].
